Class 11

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 12 Chemical Equilibrium Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 12 Chemical Equilibrium

1. Choose the correct option

Question A.
The equlilibrium , H₂O_(l) ⇌ H⁺_(aq) + OH^–_(aq) is
a. dynamic
b. static
c. physical
d. mechanical
Answer:
a. dynamic

Question B.
For the equlibrium, A ⇌ 2B + Heat, the number of ‘A’ molecules increases if
a. volume is increased
b. temperature is increased
c. catalyst is added
d. concerntration of B is decreased
Answer:
b. temperature is increased

Question C.
For the equilibrium Cl_2(g) + 2NO_(g) ⇌ 2NOCl_(g) the concerntration of NOCl will increase if the equlibrium is disturbed by ………..
a. adding Cl₂
b. removing NO
c. adding NOCl
d. removal of Cl₂
Answer:
a. adding Cl₂

Question D.
The relation between K_c and K_p for the reaction A_(g) + B_(g) ⇌ 2C_(g) + D_(g) is
a. K_c = K_p/RT
b. K_p = Kc²
c. K_c = \(\frac{1}{\sqrt{\mathrm{Kp}}}\)
d. K_p/K_c = 1
Answer:
a. K_c = K_p/RT

Question E.
When volume of the equilibrium reaction C_(s) + H₂O_(g) ⇌ CO_(g) + H_2(g) is increased at constant temperature the equilibrium will
a. shift from left to right
b. shift from right to left
c. be unaltered
d. can not be predicted
Answer:
a. shift from left to right

2. Answer the following

Question A.
State Law of Mass action.
Answer:
Law of mass action: The law of mass action states that the rate of a chemical reaction at each instant is proportional to the product of concentrations of all the reactants.

Question B.
Write an expression for equilibrium constant with respect to concerntration.
Answer:
For a reversible chemical reaction at equilibrium, aA + bB ⇌ cD + dD
Equilibrium constant (K_c) = \(\frac{[C]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}}\)

Question C.
Derive mathematically value of K_p for A_(g) + B_(g) ⇌ C_(g) + D_(g).
Answer:
When the concentrations of reactants and products in gaseous reactions are expressed in terms of their partial pressure, then the equilibrium constant is represented as K_p.
∴ For the reaction,
A_(g)+ B_(g) ⇌ C_(g) + D_(g)
the equilibrium constant (K_C) can be expressed using partial pressure as: K_p = \(\frac{P_{C} \times P_{D}}{P_{A} \times P_{B}}\)
Where P_A, P_B, P_C and P_D are equilibrium partial pressures of A, B, C and D respectively.

Question D.
Write expressions of K_C for following chemical reactions
i. 2SO_2(g) + O_2(g) ⇌ 2SO_3(g)
ii. N₂O_4(g) ⇌ 2NO_2(g)
Answer:
i. 2SO_2(g) + O_2(g) ⇌ 2SO_3(g)
K_c = \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}\)

ii. N₂O_4(g) ⇌ 2NO_2(g)
K_c = \(\frac{\left[\mathrm{NO}_{2}\right]^{2}}{\left[\mathrm{~N}_{2} \mathrm{O}_{4}\right]}\)

Question E.
Mention various applications of equilibrium constant.
Answer:
Various applications of equilibrium constant:

• Prediction of the direction of the reaction
• To know the extent of the reaction
• To calculate equilibrium concentrations
• Link between chemical equilibrium and chemical kinetics

Question F.
How does the change of pressure affect the value of equilibrium constant ?
Answer:
The change of pressure does not affect the value of equilibrium constant.

Question J.
Differentiate irreversible and reversible reaction.
Answer:
Irreversible reaction:

1. Products are not converted back to reactants.
2. Reaction stops completely and almost goes to completion.
3. It can be carried out in an open or closed vessel.
4. It takes place only in one direction. It is represented by →
5. e.g. C_(s) + O_2(g) → CO_2(g)

Reversible reaction:

1. Products arc converted back to reactants.
2. Reaction appears to have stopped but does not undergo completion.
3. It is generally carried out in a closed vessel.
4. It takes place in both directions. It is represented by ⇌
5. e.g. N_2(g) + O_2(g) ⇌ 2NO_(g)

Question K.
Write suitable conditions of concentration, temperature and pressure used during manufacture of ammonia by Haber process.
Answer:
i. Concentration: Addition of H₂ or N₂ both favours forward reaction. This increases the yield of NH₃.
ii. Temperature: The formation NH₃ is exothermic. Hence, low temperature should favour the formation of NH₃. However, at low temperatures, the rate of reaction is small. At high temperatures, the reaction occurs rapidly but decomposition of NH₃ occurs. Hence, optimum temperature of about 773 K is used.
iii. Pressure: The forward reaction is favoured with high pressure as it proceeds with decrease in number of moles. At high pressure, the catalyst becomes inefficient. Therefore, optimum pressure needs to be used. The optimum pressure is about 250 atm.

Question L.
Relate the terms reversible reactions and dynamic equilibrium.
Answer:

• Reversible reactions are the reactions which do not go to completion and occur in both the directions simultaneously.
• If such a reaction is allowed to take place for a long time, so that the concentrations of the reactants and products do not vary with time, then the reaction will attain equilibrium.
• Since, both the forward and backward reactions continue to take place in opposite directions in the same speed, the equilibrium achieved is dynamic in nature.

Thus, if the reaction is not reversible then it cannot attain dynamic equilibrium.

Question M.
For the equilibrium.
\(\mathrm{BaSO}_{4(\mathrm{~s})} \rightleftharpoons \mathrm{Ba}_{(\mathrm{aq})}^{2+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
state the effect of
a. Addition of Ba²⁺ ion.
b. Removal of SO₄^2- ion
c. Addition of BaSO_4(s)
on the equilibrium.
Answer:
a. Addition of Ba²⁺ ion will favour the reverse reaction, (that is, equilibrium shifts from right to left). This increases the amount of BaSO₄.
b. Removal of \(\mathrm{SO}_{4}^{2-}\) ion will favour the forward reaction, (that is, equilibrium shifts from left to right). This decreases the amount of BaSO₄.
c. Addition of BaSO_4(s) will not affect the equilibrium as the equilibrium constant expression does not include pure solids.

3. Explain :

Question A.
Dynamic nature of chemical equilibrium with suitable example.
Answer:
Dynamic nature of chemical equilibrium:
i. Consider a chemical reaction: A ⇌ B.
K_c = [B]/[A]
At equilibrium, the ratio of concentration of the product to that of the concentration of the reactant is constant and this is equal to K_c.

ii. At this stage reaction takes place in both the directions with same speed although the reaction appears to have stopped. Thus, the chemical equilibrium is dynamic in nature. Dynamic means moving and at a microscopic level, the system is in motion.

iii. For example, in the reaction between H₂ and I₂ to form HI, the colour of the reaction mixture becomes constant because the concentrations of H₂, I₂ and HI become constant at equilibrium.
H₂ + I₂ ⇌ 2HI
Thus, when equilibrium is reached, the reaction appears to have stopped. However, this is not the case. The reaction is still going on in the forward and backward direction but the rate of forward reaction is equal to the rate of backward reaction. Hence, chemical equilibrium is dynamic in nature and not static.

Question B.
Relation between K_c and K_p.
Answer:
Consider a general reversible reaction:
aA_(g) + bB_(g) ⇌ cC_(g) + dD_(g)
The equilibrium constant (K_p) in terms of partial pressure is given by equation:
K_p = \(\frac{\left(P_{C}\right)^{c}\left(P_{D}\right)^{d}}{\left(P_{A}\right)^{a}\left(P_{B}\right)^{b}}\) …………(1)
For a mixture of ideal gases, the partial pressure of each component is directly proportional to its concentration at constant temperature.
For component A,
P_AV = n_ART
P_A = \(\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{V}}\) × RT
\(\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{V}}\) is molar concentration of A in mol dm^-3 V
∴ P_A = [A]RT where, [A] = \(\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{V}}\)
Similarly, for other components, P_B = [B]RT, P_C = [C]RT, P_D = [D]RT
Now substituting equations for P_A, P_B, P_C, P_D in equation (1), we get

where Δn = (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced chemical equation.
R = 0.08206 L atm K^-1 mol^-1
[Note: While calculating the value of K_p, pressure should be expressed in bar, because standard state of pressure is 1 bar. 1 pascal (Pa) = 1 N m^-2 and 1 bar = 10⁵ Pa]

Question C.
State and explain Le Chatelier’s principle with reference to
1. change in temperature
2. change in concerntration.
Answer:
Statement: When a system at equilibrium is subjected to a change in any of the factors determining the equilibrium conditions of a system, system will respond in such a way as to minimize the effect of change.

1. Change in temperature:

• Consider the equilibrium reaction,
  PCl_5(g) ⇌ PCl_3(g) + Cl_2(g) + 92.5 kJ
• The forward reaction is exothermic. According to Le Chatelier’s principle an increase in temperature shifts the position of equilibrium to the left.
• The reverse reaction is endothermic. An endothermic reaction consumes heat. Therefore, the equilibrium must shift in the reverse direction to use up the added heat (heat energy converted to chemical energy).
• Thus, an increase in temperature favours formation of PCl₅ while a decrease in temperature favours decomposition of PCl₅.

2. Change in concentration:

• Consider reversible reaction representing production of ammonia (NH3).
  N_2(g) + 3H_2(g) ⇌ 2NH_3(g) + Heat
• According to Le Chatelier’s principle, when H₂ or N₂ is added to equilibrium, the effect of addition of H₂ or N₂ or is reduced by shifting the equilibrium from left to right so that the added N₂ or H₂ is consumed.
• The forward reaction occurs to a large extent than the reverse reaction until the new equilibrium is established. As a result, the yield of NH₃ is increases.
• In general, if the concentration of one of the species in equilibrium mixture is increased, the position of equilibrium shifts in the opposite so as to reduce the concentration of this species. However, the equilibrium constant remains unchanged.

Question D.
a. Reversible reaction
b. Rate of reaction
Answer:
a. Reversible reaction:
i. Reactions which do not go to completion and occur in both the directions simultaneously are called reversible reactions.
ii. Reversible reactions proceed in both directions. The direction from reactants to products is the forward reaction, whereas the opposite reaction from products to reactants is called the reverse or backward reaction.
iii. A reversible reaction is denoted by drawing in between the reactants and product a double arrow, one pointing in the forward direction and other in the reverse direction (⇌ or ⇄).
ii. At high temperature in an open container, the CO₂ gas formed will escape away. Therefore, it is not possible to obtain back
e.g. a. H_2(g) + I_2(g) ⇌ 2HI_(g)
b. CH₃COOH_(aq) + H₂O_(l) ⇌ CH₃COO^–_(aq) + H₃O⁺_(aq)

b. Rate of reaction:
Rate of a chemical reaction:
i. The rate of a chemical reaction can be determined by measuring the extent to which the concentration of a reactant decreases in the given time interval, or extent to which the concentration of a product increases in the given time interval.
ii. Mathematically, the rate of reaction is expressed as:
Rate = \(-\frac{\mathrm{d}[\text { Reactant }]}{\mathrm{dT}}=\frac{\mathrm{d}[\text { Product }]}{\mathrm{dT}}\)
where, d[reactant] and d[product] are the small decrease or increase in concentration during the small time interval dT.

Question E.
What is the effect of adding chloride on the position of the equilibrium ?
AgCl_(s) ⇌ Ag⁺_(aq) + Cl^–_(aq)
Answer:
Addition of Cl ion will favour the reverse reaction, (that is, equilibrium shift from right to left) This increases the amount of AgCl.

11th Chemistry Digest Chapter 12 Chemical Equilibrium Intext Questions and Answers

Can you recall? (Textbook Page No. 174)

Question 1.
What are the types of the following changes?
Natural waterfall, spreading of smoke from burning incense stick, diffusion of fragrance of flowers.
Answer: Natural waterfall, spreading of smoke from burning incense stick and diffusion of fragrance of flowers are irreversible physical changes.

Try this. (Textbook Page No. 174)

Question 1.
Dissolve 4 g cobalt chloride in 40 mL water. It forms a reddish pink solution. Add 60 mL concentrated HCl to this. It will turn blue. Take 5 mL of this solution in a test tube and place it in a beaker containing ice water mixture. The colour of solution will become pink. Place the same test tube in a beaker containing water at 90 °C. The colour of the solution turns blue.
Answer:
Inference: The colour change of the solution from pink to blue is caused by the chemical reaction. On changing the temperature, the direction of the reaction reverses. This indicates that the chemical reaction is reversible. This activity is an example of a reversible chemical reaction.
The reaction can be written as:

Can you tell? (Textbook Page No. 174)

Question 1.
What does violet colour of the solution in the activity mentioned in Q.2 indicate?
Answer:
In the reaction, the reactant \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) is pink in colour and the product \(\mathrm{CoCl}_{4}^{2-}\) is blue in colour. When the solution contains both the reactant and product, the resulting solution will appear violet. This indicates that the reaction has attained equilibrium (that is, the reaction proceeds in both the direction with equal rates and is a reversible reaction).

(Textbook Page No. 174)

Question 1.
Calcium earbonate when heated strongly, decomposes to form calcium oxide and carbon dioxide.
i. If this reaction is carried out in a closed container, what will we observe?
ii. Consider this reaction occurring in an open system or container, what will happen? Can we obtain back calcium carbonate?
Answer:
At high temperature in a closed container, we will find that after certain time, some calcium carbonate is present. If we continue the experiment over a longer period of time at the same temperature, the concentrations of calcium carbonate, calcium oxide and carbon dioxide remain unchanged. The reaction thus appears to have stopped and the system has attained the equilibrium. Actually, the reaction does not stop but proceeds in both the directions with equal rates. In other words, calcium carbonate decomposes to give calcium oxide and carbon dioxide at a particular rate. Exactly at the same rate the calcium oxide and carbon dioxide recombine and form calcium carbonate. Thus, in closed container, reversible reaction occurs.

ii. At high temperature in an open container, the CO₂ gas formed will escape away. Therefore, it is not possible to obtain back calcium carbonate. Thus, in an open container, irreversible reaction occurs.
\(\mathrm{CaCO}_{3(\mathrm{~s})} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_{2(\mathrm{~g})}\)

Internet my friend (Textbook Page No. 175)

Question 1.
i. Equilibrium existing in the formation of oxyhaemoglobin in human body
ii. Refrigeration system in equilibrium
Answer:
i. Equilibrium existing in the formation of oxyhaemoglobin in human body:
Oxygen is transported in the body with the assistance of red blood cells. The red blood cells contain a pigment called haemoglobin. Each haemoglobin molecule binds four oxygen molecules to form oxyhaemoglobin. Thus, the oxygen molecules are carried to individual cells in the body tissue where they are released.

The binding of oxygen to haemoglobin is a reversible reaction.
Hb + 4O₂ ⇌ Hb.4O₂
When the oxygen concentration is high (in the lungs), haemoglobin and oxygen combine to form oxyhaemoglobin and the reaction achieves equilibrium. But, when the oxygen concentration is low (in the body tissue), the reverse reaction occurs, that is, oxyhaemoglobin dissociates to haemoglobin and oxygen.
Thus, an equilibrium exists in the formation of oxyhaemoglobin in the human body.

ii. Refrigeration system in equilibrium:
a. Refrigeration system works on the principle of thermal equilibrium i.e., when a cold body comes in contact with a hot body then the heat flows from hot body to cold body until both the bodies attain the same temperature.
b. In the same way, a liquid (called as refrigerant) passes through the various compartments in the refrigerator and eventually lowers the temperature inside the refrigerator. This cycle is briefly described below:
Refrigerant flows through the compressor, which raises the pressure of the refrigerant. Next, the refrigerant flows through the condenser, where it condenses from vapor form to liquid form, giving off heat in the process. The heat given off is what j makes the condenser “hot to the touch.” After the condenser, the refrigerant goes through the expansion valve, where it experiences a pressure drop. Finally, the refrigerant goes to the evaporator. The refrigerant draws heat from the evaporator which causes the refrigerant to vaporize. The evaporator draws heat from the region that is to be cooled. The vaporized refrigerant goes back to the compressor to restart the cycle. In each of the heat transfer process, equilibrium is achieved (that is, heat given off is equivalent to the cooling achieved.)

[Note: Students are expected to collect additional information about equilibrium existing in the formation of oxyhaemoglobin in human body’ and ‘refrigeration system in equilibrium on their own.]

Try this. (Textbook Page No. 176)

Question 1.
i. Place some iodine crystals in a closed vessel. Observe the change in colour intensity in it.
ii. What do you see in the flask after some time?
Answer:
i. The vessel gets slowly filled up with violet coloured vapour of iodine. After a certain time, the intensity of violet colour becomes stable.
ii. After sometime, both solid iodine and iodine vapour are present in the closed vessel. Iodine crystals will be seen deposited near the mouth of the flask and violet coloured vapour will be filled in the entire flask. It means solid iodine sublimes to give iodine vapour and the iodine vapour condenses to form solid iodine. The stable intensity of the colour indicates a state of equilibrium between solid and vapour iodine.

Try this. (Textbook Page No. 176)

Question 1.
i. Dissolve a given amount of sugar in minimum amount of water at room temperature.
ii. Increase the temperature and dissolve more amount of sugar in the same amount of water to make a thick sugar syrup solution.
iii. Cool the syrup to the room temperature.
Answer:
Observation: Sugar crystals separate out.
Inference: The sugar syrup solution prepared is a saturated solution. Therefore, additional amount of sugar cannot be dissolved in it at room temperature.
In a saturated solution, there exists dynamic equilibrium between the solute molecules in the solid state and in dissolved state.
Sugar_(aq) ⇌ Sugar_(s)
The rate of dissolution of sugar = The rate of crystallization of sugar.
However, when it is heated, additional amount of sugar can be dissolved in it. But when such a thick sugar syrup is cooled again to room temperature, sugar crystals separate out.

Do you know? (Textbook Page No. 177)

Question 1.
What is a saturated solution?
Answer:
A saturated solution is the solution when additional solute cannot be dissolved in it at the given temperature. The concentration of solute in a saturated solution depends on temperature.

Observe and discuss. (Textbook Page No. 177)

Question 1.
Colourless N₂O₄ taken in a closed flask is converted to NO₂ (a reddish brown gas).

Answer:
Observation: Initially, the colourless gas (N₂O₄) turns to reddish brown (NO₂) gas. After sometime, the colour becomes lighter indicating the formation of N₂O₄ from NO₂.
Inference: This indicates that the reaction is reversible. In such reaction, the reactants combine to form the products and the products combine to give the reactants. As soon as the forward reaction produces any NO₂, the reverse reaction begins and NO₂, starts combining back to N₂O₄. At equilibrium, the concentrations of N₂O₄ and NO₂ remain unchanged and do not vary with time, because the rate of formation of NO₂ is equal to the rate of formation of N₂O₄.

[Note: For any reversible reaction in a closed system whenever the opposing reactions (forward and reaction) are occurring at different rates, the forward reaction will gradually become slower and the reverse reaction will become faster. Finally, the rates become equal and equilibrium is established.]

Discuss (Textbook Page No. 177)

i. Consider the following dissociation reaction:

The reaction is carried out in a closed vessel starting with hydrogen iodide.
ii. Now, let us start with hydrogen and iodine vapour in a closed container at a certain temperature.
H_2(g) + I_2(g) ⇌ 2HI_(g)
Answer:
i. Starting with hydrogen iodide:
Observations:
a. At first, there is an increase in the intensity of violet colour.
b. After certain time, the increase in the intensity of violet colour stops.
c. When contents in a closed vessel are analyzed at this stage, it is observed that reaction mixture contains the hydrogen iodide, hydrogen and iodine with their concentrations being constant over time.
Inference:
The rate of decomposition of HI becomes equal to the rate of combination of H₂ and I₂. At equilibrium, no net change is observed and both reactions continue to occur at equal rates.
Thus, the reaction represents chemical equilibrium.

ii. Starting with hydrogen and iodine:
Observations:
a. At first, there is a decrease in the intensity of violet colour.
b. After certain time, the decrease in the intensity of violet colour stops.
c. When contents in a closed vessel are analyzed at this stage, it is observed that reaction mixture contains hydrogen, iodine and hydrogen iodide with their concentrations being constant over time.
Inference:
The rate of combination of H₂ and I₂ becomes equal to the rate of decomposition of HI. The reaction attains chemical equilibrium.

Can you recall? (Textbook Page No. 180)

Question 1.
Write ideal gas equation with significance of each term involved in it.
Answer:
Ideal gas equation is PV = nRT.
where, P = Pressure of the gas
V = Volume of the gas
n = Number of moles of the gas
R = universal gas constant
T = Absolute temperature of the gas

Just think. (Textbook page no. 181)

Question 1.
Two processes, which are taking place in opposite directions are in equilibrium. How to write equilibrium constant expersions for heterogeneous equilibrium?
Answer:
Equilibrium in a system having more than one phase is called heterogeneous equilibrium.
If ethanol is placed in a conical flask, liquid-vapour equilibrium is established.
C₂H₅OH_(l) ⇌ C₂H₅OH_(g)
For a given temperature,
K_c = \(\frac{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}_{(g)}\right]}{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}_{(l)}\right]}\)
But [C₂H₅OH_(l)] = 1
∴ K_c = [C₂H₅OH_(g)]
Thus, at any given temperature, density is constant irrespective of the amount of liquid, and the term in the denominator is also constant.
ii. similarly, consider I_2(g) ⇌ I_2(g)
K_c = [I₂(g)]
iii. Thus, the expression for equilibrium constant does not contain the concentration of pure solids and pure liquids. That is because for any pure liquid and solid, the concentration is simply its density and this will not change no matter how much solid or liquid is used. Hence, the expression for heterogeneous equilibrium only uses the concentration of gases and dissolved substances (aq.). Solids are pure substances with unchanging concentrations and thus equilibria including solids are simplified.

Can you tell? (Textbook Page No. 183)

Question 1.
Comment on the extent to which the forward reaction will proceed, from the magnitude of the equilibrium constant for the following reactions:
i. H_2(g) + I_2(g) ⇌ 2HI_(g), K_c = 20 at 550 K
ii. H_2(g) + Cl_2(g) ⇌ 2HCl_(g), K_c = 1018 at 550 K
Answer:
i. For the reaction, K_c = 20 at 550 K
If the value of K_c is the range of 10^-3 to 10³, the forward and reverse proceed to equal extents.
Hence, the given reaction will form appreciable concentrations of both reactants and the product at equilibrium.

ii. For the reaction, K_c = 10¹⁸ at 550 K
If the value of K_c >>> 10³, forward reaction is favoured.
Hence, the given reaction will proceed in the forward direction and will nearly go to completion.

Use your brain power (Textbook Page No. 183)

Question 1.
The value of Kc for the dissociation reaction:
H_2(g) ⇌ 2H_(g) is 1.2 × 10^-42 at 500 K.
Does the equilibrium mixture contain mainly hydrogen molecules or hydrogen atoms?
Answer:
When the value of K_c is very low (that is, K_c < 10^-3), then at equilibrium, only a small fraction of the reactants is converted into products.
For the given reaction, K_c <<< 10³ at 500 K.
Hence, the equilibrium mixture contains mainly hydrogen molecules.

Internet my friend (Textbook Page No. 183)

Question 1.
Collect information about chemical equilibrium.
Answer:
https://www.chemguide.co.uk/physical/equilibria/introduction.html
[Note: Students can use the above link as reference and collect information about chemical equilibrium.]

Can you tell? (Textbook Page No. 188)

i. If NH₃ is added to the equilibrium system (Haber process), in which direction will the equilibrium shift to consume added NH₃ to reduce the effect of stress?
ii. In this process, out of the reactions (reverse and forward reaction), which reaction will occur to a greater extent?
iii. What will be the effect on yield of NH₃?
Answer:
i. If NH₃ is added to the equilibrium system, the equilibrium will shift from right to left to consume added NH₃ to reduce the effect of stress.
ii. If NH₃ is added to the equilibrium system, then reverse reaction will occur to greater extent.
iii. If NH₃ is added to the equilibrium system, the equilibrium will shift in reverse direction and the yield of NH₃ will decrease.

Internet my friend (Textbook Page No. 188)

i. Collect information about Haber process in chemical equilibrium.
ii. Youtube.Freescienceslessons: The Haber process
Answer:
i. https://www.chemguide.co.uk/physical/equilibria/haber.html
[Note: Students can use the above link as reference and collect information about chemical equilibrium involved in Haber process.]
ii. Students are expected to refer ‘The Haber process ’ on YouTube channel ‘Freescienceslessons’

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 12 Magnetism Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 12 Magnetism

1. Choose the correct option.

Question 1.
Let r be the distance of a point on the axis of a bar magnet from its center. The magnetic field at r is always proportional to
(A) \(\frac {1}{r^2}\)
(B) \(\frac {1}{r^3}\)
(C) \(\frac {1}{r}\)
(D) Not necessarily \(\frac {1}{r^3}\) at all points
Answer:
(B) \(\frac {1}{r^3}\)

Question 2.
Magnetic meridian is the plane
(A) perpendicular to the magnetic axis of Earth
(B) perpendicular to geographic axis of Earth
(C) passing through the magnetic axis of Earth
(D) passing through the geographic axis
Answer:
(C) passing through the magnetic axis of Earth

Question 3.
The horizontal and vertical component of magnetic field of Earth are same at some place on the surface of Earth. The magnetic dip angle at this place will be
(A) 30°
(B) 45°
(C) 0°
(D) 90°
Answer:
(B) 45°

Question 4.
Inside a bar magnet, the magnetic field lines
(A) are not present
(B) are parallel to the cross sectional area of the magnet
(C) are in the direction from N pole to S pole
(D) are in the direction from S pole to N pole
Answer:
(D) are in the direction from S pole to N pole

Question 5.
A place where the vertical components of Earth’s magnetic field is zero has the angle of dip equal to
(A) 0°
(B) 45°
(C) 60°
(D) 90°
Answer:
(A) 0°

Question 6.
A place where the horizontal component of Earth’s magnetic field is zero lies at
(A) geographic equator
(B) geomagnetic equator
(C) one of the geographic poles
(D) one of the geomagnetic poles
Answer:
(D) one of the geomagnetic poles

Question 7.
A magnetic needle kept nonparallel to the magnetic field in a nonuniform magnetic field experiences
(A) a force but not a torque
(B) a torque but not a force
(C) both a force and a torque
(D) neither force nor a torque
Answer:
(C) both a force and a torque

2. Answer the following questions in brief.

Question 1.
What happens if a bar magnet is cut into two pieces transverse to its length/ along its length?
Answer:
i. When a magnet is cut into two pieces, then each piece behaves like an independent magnet.

ii. When a bar magnet is cut transverse to its length, the two pieces generated will behave as independent magnets of reduced magnetic length. However, the pole strength of all the four poles formed will be same as that of the original bar magnet. Thus, the new dipole moment of the smaller magnets will be,

iii. When the bar magnet is cut along its length, the two pieces generated will behave like an independent magnet with reduced pole strength. However, the magnetic length of both the new magnets will be same as that of the original bar magnet. Thus, the new dipole moment of the smaller magnets will be,

Question 2.
What could be the equation for Gauss’ law of magnetism, if a monopole of pole strength p is enclosed by a surface?
Answer:
i. According to Gauss’ law of electrostatics, the net electric flux through any Gaussian surface is proportional to net charge enclosed in it. The equation is given as,
ø_E = ∫\(\vec{E}\) . \(\vec{dS}\) = \(\frac {q}{ε_0}\)

ii. Similarly, if a monopole of a magnet of pole strength p exists, the Gauss’ law of magnetism in S.I. units will be given as,
ø_E = ∫\(\vec{B}\) . \(\vec{dS}\) = µ₀P

3. Answer the following questions in detail.

Question 1.
Explain the Gauss’ law for magnetic fields.
Answer:
i. Analogous to the Gauss’ law for electric field, the Gauss’ law for magnetism states that, the net magnetic flux (ø_B) through a closed Gaussian surface is zero. ø_B = ∫\(\vec{B}\) . \(\vec{dS}\) = 0

ii. Consider a bar magnet, a current carrying solenoid and an electric dipole. The magnetic field lines of these three are as shown in figures.

iii. The areas (P) and (Q) are the cross – sections of three dimensional closed Gaussian surfaces. The Gaussian surface (P) does not include poles while the Gaussian surface (Q) includes N-pole of bar magnet, solenoid and the positive charge in case of electric dipole.

iv. The number of lines of force entering the surface (P) is equal to the number of lines of force leaving the surface. This can be observed in all the three cases.

v. However, Gaussian surface (Q) of bar magnet, enclose north pole. As, even thin slice of a bar magnet will have both north and south poles associated with it, the number of lines of Force entering surface (Q) are equal to the number of lines of force leaving the surface.

vi. For an electric dipole, the field lines begin from positive charge and end on negative charge. For a closed surface (Q), there is a net outward flux since it does include a net (positive) charge.

vii. Thus, according to the Gauss’ law of electrostatics ø_E = ∫\(\vec{E}\) . \(\vec{dS}\) = \(\frac {q}{ε_0}\), where q is the positive charge enclosed.

viii. The situation is entirely different from magnetic lines of force. Gauss’ law of magnetism can be written as ø_B = ∫\(\vec{B}\) . \(\vec{dS}\) = 0
From this, one can conclude that for electrostatics, an isolated electric charge exists but an isolated magnetic pole does not exist.

Question 2.
What is a geographic meridian? How does the declination vary with latitude? Where is it minimum?
Answer:
A plane perpendicular to the surface of the Earth (vertical plane) and passing through geographic axis is geographic meridian.

i. Angle between the geographic and the magnetic meridian at a place is called magnetic declination (a).
ii. Magnetic declination varies with location and over time. As one moves away from the true north the declination changes depending on the latitude as well as longitude of the place. By convention, declination is positive when magnetic north is east of true north, and negative when it is to the west. The declination is small in India. It is 0° 58′ west at Mumbai and 0° 41′ east at Delhi.

Question 3.
Define the angle of dip. What happens to angle of dip as we move towards magnetic pole from magnetic equator?
Answer:
Angle made by the direction of resultant magnetic field with the horizontal at a place is inclination or angle of dip (ø) at the place.
At the magnetic pole value of ø = 90° and it goes on decreasing when we move towards equator such that at equator value of (ø) = 0°.

4. Solve the following problems.

Question 1.
A magnetic pole of bar magnet with pole strength of 100 Am is 20 cm away from the centre of a bar magnet. Bar magnet has pole strength of 200 Am and has a length 5 cm. If the magnetic pole is on the axis of the bar magnet, find the force on the magnetic pole.
Answer:
Given that, (q_m)₁ = 200 Am
and (2l) = 5 cm = 5 × 10^-2 m
∴ m = 200 × 5 × 10^-2 = 10 Am²
For a bar magnet, magnetic dipole moment is,
m = q_m (21)
For a point on the axis of a bar magnet at distance, r = 20 cm = 0.2 m,
B_a = \(\frac{\mu_{0}}{4 \pi} \times \frac{2 m}{r^{3}}\)
= 10^-7 × \(\frac{2 \times 10}{(0.2)^{3}}\)
= 0.25 × 10^-3
= 2.5 × 10^-4 Wb/m²
The force acting on the pole will be given by,
F = q_m B_a = 100 × 2.5 × 10^-4
= 2.5 × 10^-2 N

Question 2.
A magnet makes an angle of 45° with the horizontal in a plane making an angle of 30° with the magnetic meridian. Find the true value of the dip angle at the place.
Answer:
Let true value of dip be ø. When the magnet is kept 45° aligned with declination 30°, the horizontal component of Earth’s magnetic field.
B’_H = B_H cos 30° Whereas, vertical component remains unchanged.
∴ For apparent dip of 45°,
tan 45° = \(\frac{\mathrm{B}_{\mathrm{V}}^{\prime}}{\mathrm{B}_{\mathrm{H}}^{\prime}}=\frac{\mathrm{B}_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}} \cos 30^{\circ}}=\frac{\mathrm{B}_{\mathrm{v}}}{\mathrm{B}_{\mathrm{H}}} \times \frac{1}{\cos 30^{\circ}}\)
But, real value of dip is,
tan ø = \(\frac {B_V}{B_H}\)
∴ tan 45° = \(\frac {tan ø}{cos 30°}\)
∴ tan ø = tan 45° × cos 30°
= 1 × \(\frac {√3}{2}\)
∴ ø = tan^-1 (0.866)

Question 3.
Two small and similar bar magnets have magnetic dipole moment of 1.0 Am² each. They are kept in a plane in such a way that their axes are perpendicular to each other. A line drawn through the axis of one magnet passes through the centre of other magnet. If the distance between their centres is 2 m, find the magnitude of magnetic field at the midpoint of the line joining their centres.
Answer:
Let P be the midpoint of the line joining the centres of two bar magnets. As shown in figure, P is at the axis of one bar magnet and at the equator of another bar magnet. Thus, the magnetic field on the axis of the first bar magnet at distance of 1 m from the centre will be,

B_a = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}\)
= 10^-7 × \(\frac {2×1.0}{(1)^3}\)
= 2 × 10^-7 Wb/m²
Magnetic field on the equator of second bar magnet will be,
B_eq = \(\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}\)
= 10^-7 × \(\frac {1.0}{(1)^3}\)
= 1 × 10^-7 Wb/m²
The net magnetic field at P,
B_net = \(\sqrt {B_a^2+B_{eq}^2}\)
= \(\sqrt {(2×10^{-7})^2+(1×10^{-7})^2}\)
= \(\sqrt {(10^{-7})^2×(4+1)}\)
= √5 × 10^-7 Wb/m²

Question 4.
A circular magnet is made with its north pole at the centre, separated from the surrounding circular south pole by an air gap. Draw the magnetic field lines in the gap. Draw a diagram to illustrate the magnetic lines of force between the south poles of two such magnets.
Answer:
i. For a circular magnet:

Question 5.
Two bar magnets are placed on a horizontal surface. Draw magnetic lines around them. Mark the position of any neutral points (points where there is no resultant magnetic field) on your diagram.
Answer:
The magnetic lines of force between two magnets will depend on their relative positions. Considering the magnets to be placed one besides the other as shown in figure, the magnetic lines of force will be as shown.

11th Physics Digest Chapter 12 Magnetism Intext Questions and Answers

Can you recall? (Textbook page no. 221)

Question 1.
What are the magnetic lines of force?
Answer:
The magnetic field around a magnet is shown by lines going from one end of the magnet to the other. These lines are named as magnetic lines of force.

Question 2.
What are the rules concerning the lines of force?
Answer:
i. Magnetic lines of force originate from the north pole and end at the south pole.
ii. The magnetic lines of force of a magnet or a solenoid form closed loops. This is in contrast to the case of an electric dipole, where the electric lines of force originate from the positive charge and end on the negative charge.
iii. The direction of the net magnetic field \(\vec {B}\) at a point is given by the tangent to the magnetic line of force at that point.
iv. The number of lines of force crossing per unit area decides the magnitude of magnetic field \(\vec {B}\)
v. The magnetic lines of force do not intersect. This is because had they intersected, the direction of magnetic field would not be unique at that point.

Question 3.
What is a bar magnet?
Answer:
Bar magnet is a magnet in the shape of bar having two poles of equal and opposite pole strengths separated by certain distance (2l).

Question 4.
If you freely hang a bar magnet horizontally, in which direction will it become stable?
Answer:
A bar magnet suspended freely in air always aligns itself along geographic N-S direction.

Try this (Textbook page no. 221)

You can take a bar magnet and a small compass needle. Place the bar magnet at a fixed position on a paper and place the needle at various positions. Noting the orientation of the needle, the magnetic field direction at various locations can be traced.
Answer:
When a small compass needle is kept at any position near a bar magnet, the needle always aligns itself in the direction parallel to the direction of magnetic lines of force.

Hence, by placing it at different positions, A, B, C, D,… as shown in the figure, the direction of magnetic lines of force can be traced. The direction of magnetic field will be a tangent at that point.

Internet my friend: (Text book page no. 227)

https://www.ngdc.noaa.gov
[Students are expected to visit above mentioned link and collect more information about Geomagnetism.]

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.4

Question 1.
Find the slope, x-intercept, y-intercept of each of the following lines, i. 2x + 3y-6 = 0 ii. 3x-y-9 = 0 iii. x + 2y = 0
Solution:
i. Given equation of the line is 2x + 3y – 6 = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 2, b = 3, c = -6

ii. Given equation of the line is 3x – y – 9 = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 3, b = – 1, c = – 9

iii. Given equation of the line is x + 2y = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 1, b = 2, c = 0

Question 2.
Write each of the following equations in ax + by + c = 0 form.
i. y = 2x – 4
ii. y = 4
iii. \(\frac{x}{2}+\frac{y}{4}=1\)
iv. \(\frac{x}{3}-\frac{y}{2}=0\)
i. y = 2x – 4
∴ 2x – y – 4 = 0 is the equation in ax + by + c = 0 form.

ii. y = 4
∴ 0x + 1y – 4 = 0 is the equation in ax + by + c = 0 form.

iii. \(\frac{x}{2}+\frac{y}{4}=1\)
∴ \(\frac{2 x+y}{4}\)
∴ 2x + y – 4 = 0 is the equation in ax + by + c = 0 form.

iv. \(\frac{x}{3}-\frac{y}{2}=0\)
∴ 2x – 3y = 0
∴ 2x – 3y + 0 = 0 is the equation in ax + by + c = 0 form.
[Note: Answer given in the textbook is ‘2x – 3y – 6 = 0’. However, as per our calculation it is ‘2x-3y + 0 = 0’.]

Question 3.
Show that the lines x – 2y – 7 = 0 and 2x – 4y + 15 = 0 are parallel to each other.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x – 4y + 15 = 0.
∴ m₂ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{-4}=\frac{1}{2}\)
Since m₁ = m₂
the given lines are parallel to each other.

Question 4.
Show that the lines x – 2y – 7 = 0 and 2x + y + 1 = 0 are perpendicular to each other. Find their point of intersection.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x + y + 1 = 0.
∴ m₂ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{1}=-2\)
Since m₁ x m₁ = \(\frac{1}{2}\) x (- 2) = -1,
the given lines are perpendicular to each other. Consider,
x – 2y – 7 = 0 …(i)
2x + y + 1 =0 …(ii)
Multiplying equation (ii) by 2, we get
4x + 2y + 2 = 0 …(iii)
Adding equations (i) and (iii), we get
5x – 5 = 0
∴ x = 1
Substituting x = 1 in equation (ii), we get
2 + y + 1 = 0
∴ y = – 3
∴ The point of intersection of the given lines is (1,-3).

Question 5.
If the line 3x + 4y = p makes a triangle of area 24 square units with the co-ordinate axes, then find the value of p.
Solution:
Let the line 3x + 4y = p cuts the X and Y axes at points A and B respectively.
3x + 4y = p

This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\),
where a = \(\frac{p}{3}\) and b = \(\frac{p}{4}\)
∴ A (a, 0) ≡ (\(\frac{p}{3}\), 0) and B ≡ (0, b) = (0, \(\frac{p}{4}\))
∴ OA = \(\frac{p}{3}\) and OB = \(\frac{p}{4}\)
Given, A (∆OAB) = 24 sq. units
∴ \(\left|\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\right|=24\)
∴ \(\left|\frac{1}{2} \times \frac{\mathrm{p}}{3} \times \frac{\mathrm{p}}{4}\right|=24\)
∴ p² = 576
∴ p = ± 24

Question 6.
Find the co-ordinates of the foot of the perpendicular drawn from the point A(- 2,3) to the line 3x-y -1 = 0.
Solution:
Let M be the foot of perpendicular drawn from
point A(- 2,3) to the line
3x-y- 1 = 0 …(i)
Slope of the line 3x-y – 1 = 0 is \(\frac{-3}{-1}\) =3.
Since AM ⊥ to line (i),
slope of AM = \(\frac{-1}{3}\)
∴ Equation of AM is
y – 3 = \(\frac{-1}{3}\)(x + 2)

∴ 3(y – 3) = – 1(x + 2)
∴ 3y – 9 = -x – 2
∴ x + 3y – 7 = 0 …………(ii)
The foot of perpendicular i.e., point M, is the point of intersection of equations (i) and (ii).
By (i) x 3 + (ii), we get 10x -10 = 0
∴ x = 1
Substituting x = 1 in (ii), we get
1 + 3y – 7 = 0
∴ 3y = 6
∴ y = 2
∴ The co-ordinates of the foot of the perpendicular Mare (1,2).

Question 7.
Find the co-ordinates of the circumcentre of the triangle whose vertices are A(- 2, 3), B(6, -1), C(4,3),e
Solution:
Here, A(-2, 3), B(6, -1), C(4, 3) are the vertices of ∆ABC.
Let F be the circumcentre of AABC.
Let FD and FE be the perpendicular bisectors of the sides BC and AC respectively.
∴ D and E are the midpoints of side BC and AC respectively.

Since FD passes through (5, 1) and has slope 1/2 equation of FD is
y – 1 = \(\frac{1}{2}\)(x-5)
∴ 2 (y – 1) = x – 5
∴ 2y – 2 = x – 5
∴ x – 2y – 3 = 0 …(i)
Since both the points A and C have same y co-ordinates i.e. 3,
the given points lie on the line y = 3.
Since the equation FE passes through E(1, 3),
the equation of FE is x = 1. .. .(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of x in (i), we get
1 – 2y -3 = 0
∴ y = -1
∴ Co-ordinates of circumcentre F ≡ (1, – 1).

Question 8.
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(3, – 2), B(7,6), C (-1,2).
Solution:
Let O be the orthocentre of ∆ABC.
Let AD and BE be the altitudes on the sides BC and AC respectively.

Slope of side BC = \(\frac{2-6}{-1-7}=\frac{-4}{-8}=\frac{1}{2}\)
∴ Slope of AD = – 2 [∵ AD ⊥ BC]
∴ Equation of line AD is
y – (-2) = (- 2) (x – 3)
∴ y + 2 = -2x + 6
∴ 2x + y -4 = 0 …(i)
Slope of side AC = \(\frac{-2-2}{3-(-1)}=\frac{-4}{4}\) = -1
∴ Slope of BE = 1 …[ ∵ BE ⊥ AC]
∴ Equation of line BE is
y – 6 = 1(x – 7)
∴ y – 6 = x – 1
∴ x = y + 1 …(ii)
Substituting x = y + 1 in (i), we get
2(y + 1) + y – 4 = 0
∴ 2y + 2 + y – 4 = 0
∴ 3y – 2 = 0
∴ y = \(\frac{2}{3} in (ii), we get
Substituting y = [latex]\frac{2}{3}\) in (ii), we get
x = \(\frac{2}{3}+1=\frac{5}{3}\)
∴ Co-ordinates of orthocentre, O = \(\left(\frac{5}{3}, \frac{2}{3}\right)\)

Question 9.
Show that the lines 3 – 4y + 5 = 0, lx – 8y + 5 = 0 and 4JC + 5y – 45 = 0 are concurrent. Find their point of concurrence.
Solution:
The number of lines intersecting at a point are called concurrent lines and their point of intersection is called the point of concurrence. Equations of the given lines are
3x – 4y + 5 = 0 …(i)
7x-8y + 5 = 0 …(ii)
4x + 5y – 45 = 0 …(iii)
By (i) x 2 – (ii), we get
– x + 5 = 0
∴ x = 5
Substituting x = 5 in (i), we get
3(5) – 4y + 5 = 0
∴ -4y = – 20
∴ y = 5
∴ The point of intersection of lines (i) and (ii) is given by (5, 5).
Substituting x = 5 and y = 5 in L.H.S. of (iii), we get
L.H.S. = 4(5) + 5(5) – 45
= 20 + 25 – 45
= 0
= R.H.S.
∴ Line (iii) also passes through (5, 5).
Hence, the given three lines are concurrent and the point of concurrence is (5, 5).

Question 10.
Find the equation of the line whose x-intercept is 3 and which ¡s perpendicular to the line 3x – y + 23 = 0.
Solution:
Slope of the line 3x – y + 23 = 0 is 3.
∴ Slope of the required line perpendicular to
3x – y + 23 = 0 is \(\frac{-1}{3}\)
Since the x-intercept of the required line is 3, it passes through (3, 0).
∴ The equation of the required line is ‘
y – 0 = \(\frac{-1}{3}\)(x – 3)
∴ 3y = x + 3
∴ x + 3y = 3

Question 11.
Find the distance of the origin from the line 7x + 24y – 50 = 0.
Solution:
Let p be the perpendicular distance of origin
fromtheline7x + 24y – 50 = 0
Here, a = 7, b = 24, c = -50

Question 12.
Find the distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0.
Solution:
Let p be the perpendicular distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0
Here, a = 12, b = – 5, c = – 13, x₁ = -2, y₁ = 3

Question 13.
Find the distance between parallel lines 4x – 3y + 5 = 0 and 4xr – 3y + 7 = 0.
Solution:
Equations of the given parallel lines are 4x – 3y + 5 = 0 and 4x – 3y + 1 = 0
Here, a = 4, b = – 3, c₁ = 5 and c₂ = 7
∴ Distance between the parallel lines

Question 14.
Find the distance between the parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Solution:
Equations of the given parallel lines are 3x + 2y + 6 = 0 and
9x + 6y – 1 = 0 i.e., 3x + 2y – \(\frac{7}{3}\) =0
Here, a = 3, b = 2, c₁ = 6 and c₂ = \(\frac{-7}{3}\)
∴ Distance between the parallel lines

Question 15.
Find the points on the line x + y – 4 = 0 which are at a unit distance from the line 4JC + 3y = 10.
Solution:
Let P(x₁, y₁) be a point on the line x + y – 4 = o.
∴ x₁ + y₁ – 4 = 0
∴ y₁ = 4 – x₁ …(i)
Also, distance of P from the line 4x + 3y- 10 = 0 is 1

∴ 5 = | x₁ + 2 |
∴ x₁ + 2 = ± 5
∴ x₁ + 2 = 5 or x₁ + 2 = – 5
∴ x₁ = 3 or x₁ = – 7
From (i), when x₁ = 3, y₁ = 1
and when x₁ = -7, y₁ = 11
∴ The required points are (3, 1) and (-7, 11).
[Note: The question has been modified]

Question 16.
Find the equation of the line parallel to the X-axis and passing through the point of intersection of lines x + y – 2 = 0 and 4x + 3y = 10.
Solution:
Let u = x + y – 2 = 0 and v = 4x + 3y – 10 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
∴ (x + y – 2) + k(4x + 3y – 10) = 0 …(i)
∴ x + y – 2 + 4kx + 3ky – 10k = 0
∴ x + 4kx + y + 3ky – 2 – 10k = 0
∴ (1+ 4k)x + (1 + 3k)y – 2 – 10k = 0
But, this line is parallel to X-axis.
∴ Its slope = 0
∴ \(\frac{-(1+4 k)}{1+3 k}\) = 0
∴ 1 + 4k = 0
∴ k = \(\frac{-1}{4}\)
Substituting the value of k in (i), we get
(x + y – 2) + (4x + 3y – 10) = 0
∴ 4(x +y – 2) – (4x + 3y -10 ) = 0
∴ 4x + 4y – 8 – 4x – 3y + 10 = 0
∴ y + 2 = 0, which is the equation of the required line.
[Note: Answer given in the textbook is 5y – 8= 0. However, as per our calculation it is y + 2 = 0.]

Question 17.
Find the equation of the line passing through the point of intersection of lines x + y – 2 = 0 and 2xr – 3y + 4 = 0 and making intercept 3 on the X-axis.
Solution:
Let u ≡ x + y – 2 = 0 and v ≡ 2x – 3y + 4 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
∴ (x +y – 2) + k(2x – 3y + 4) = 0 …(i)
But, x-intercept of line is 3.
∴ It passes through (3, 0).
Substituting x = 3 and y = 0 in (i), we get
(3 + 0 – 2) + k(6 – 0 + 4) = 0
∴ 1 + 10k = 0
k = \(\frac{-1}{10}\)
Substituting the value of k in (i), we get (x + y – 2) + \(\left(\frac{-1}{10}\right)\) (2x – 3y + 4) = 0
∴ 10(x + y – 2) – (2x – 3y + 4) = 0
∴ 10x + 10y -20 — 2x + 3y-4 = 0
∴ 8x + 13y – 24 = 0, which is the equation of the required line.

Question 18.
If A(4, 3), B(0, 0) and C(2, 3) are the vertices of ΔABC, then find the equation of bisector of angle BAC.
Solution:
Let the bisector of ∠ BAC meets BC at point D.
∴ Point D divides seg BC in the ratio l(AB) : l(AC)

∴ 18 (y – 3) = 6 (x – 4)
∴ 3(y – 3) = x – 4
∴ 3y – 9 = x – 4
∴ x – 3y + 5 = 0

Question 19.
D(- 1, 8), E(4, – 2), F(- 5, – 3) are midpoints of sides BC, CA and AB of AABC. Find
i. equations of sides of ΔABC.
ii. co-ordinates of the circumcentre of ΔABC.
Solution:
Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of ΔABC.
Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ΔABC.

∴ x₁ + x₂ = -10 …………. (v)
and y₁ + y₂ = – 6 …………(vi)
For x-coordinates:
Adding (i), (iii) and (v), we get
2x₁ + 2x₂ + 2x₃ = – 4
∴ x₁ + x₂ + x₃ = -2 …………..(vii)
Solving (i) and (vii), we get x₁ = 0
Solving (iii) and (vii), we get x₂ = – 10
Solving (v) and (vii), we get x₃ = 8

For y-coordinates:
Adding (ii), (iv) and (vi), we get 2y₁ + 2y₂ + 2y₃ = 6
y₁ + y₂ + y₃ = 3 …….(viii)
Solving (ii) and (viii), we get y₁ = -13
Solving (iv) and (viii), we get y₂ = 7
Solving (vi) and (viii), we get y₃ = 9
∴ Vertices of AABC are A(0, – 13), B(- 10, 7), C(8, 9)

∴ 8(y + 13) = 22x
∴ 4(y + 13) = 11x
∴ 11x – 4y – 52 = 0

ii. Here, A(0, – 13), B(- 10, 7), C(8, 9) are the vertices of ΔABC.
Let F be the circumcentre of AABC.
Let FD and FE be perpendicular bisectors of the sides BC and AC respectively.
D and E are the midpoints of side BC and AC.

∴ Slope of FD = -9 … [ ∵ FD ⊥ BC]
Since FD passes through (-1, 8) and has slope -9, equation of FD is
y – 8 = -9 (x +1)
∴ y – 8 = -9x – 9
∴ y = -9x – 1
Also, slope of AC = \(\frac{-13-9}{0-8}=\frac{11}{4}\)
∴ Slope of FE = \(\frac{-4}{11}\) [ ∵ FE ⊥ AC]
Since FE passes through (4, -2) and has slope -4
\(\frac{-4}{11}\), equation of FE is
(y + 2) = \(\frac{-4}{11}\) (x – 4)
∴ 11(y + 2) = -4(x – 4)
∴ 11y + 22 = – 4x + 16
∴ 4x + 11y = -6 …………(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value ofy in (ii), we get
4x + 11(-9x- 1) = – 6
∴ 4x – 99x -11 = – 6
∴ -95x = 5
∴ x = \(\frac{-1}{19}\)
Substituting the value of x in (i), we get
y = -9(\(\frac{-1}{19}\)) – 1 = \(\frac{-10}{19}\)
∴ Co-ordinates of circumcentre F ≡ \(\left(\frac{-1}{19}, \frac{-10}{19}\right)\)

Question 20.
0(0, 0), A(6, 0) and B(0, 8) are vertices of a triangle. Find the co-ordinates of the incentre of ∆OAB.
Solution:
Let bisector of ∠O meet AB at point D and bisector of ∠A meet BO at point E
∴ Point D divides seg AB in the ratio l(OA): l(OB)
and point E divides seg BO in the ratio l(AB): l(AO)
Let I be the incentre of ∠OAB.

∴ Point D divides AB internally in 6 : 8
i.e. 3 :4

∴ y = x …(i)
Now, by distance formula,
l(AB) = \(\begin{aligned}
&=\sqrt{(6-0)^{2}+(0-8)^{2}} \\
&=\sqrt{36+64}=10
\end{aligned}\)
l(AO) = \(\sqrt{(6-0)^{2}+(0-0)^{2}}\) = 6
∴ Point E divides BO internally in 10 : 6 i.e. 5:3

∴ -2y = x – 6
∴ x + 2y = 6 …(ii)
To find co-ordinates of incentre, we have to solve equations (i) and (ii).
Substituting y = x in (ii), we get
x + 2x = 6
∴ x = 2
Substituting the value of x in (i), we get
y = 2
∴ Co-ordinates of incentre I ≡ (2, 2).

Alternate Method:
Let I be the incentre.
I lies in the 1st quadrant.
OPIR is a square having side length r.
Since OA = 6, OP = r,
PA = 6 – r
Since PA = AQ,
AQ = 6 – r …(i)
Since OB = 8, OR = r,
BR = 8 – r
∴ BR = BQ
∴ BQ = 8 – r …(ii)

AB = BQ + AQ
Also, AB = \(\begin{aligned}
&=\sqrt{\mathrm{OA}^{2}+\mathrm{OB}^{2}} \\
&=\sqrt{6^{2}+8^{2}} \\
&=\sqrt{100}=10
\end{aligned}\)
∴ BQ + AQ= 10
∴ (8 – r) + (6 – r) = 10
∴ 2r = 14- 10 = 4
∴ r = 2
∴ I = (2,2)

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.3

Question 1.
Write the equation of the line:
i. parallel to the X-axis and at a distance of 5 units from it and above it.
ii. parallel to the Y-axis and at a distance of 5 units from it and to the left of it.
iii. parallel to the X-axis and at a distance of 4 units from the point (- 2,3).
Solution:
i. Equation of a line parallel to X-axis is y = k. Since the line is at a distance of 5 units above the X-axis, k = 5
∴ The equation of the required line is y = 5.

ii. Equation of a line parallel to the Y-axis is x = h. Since the line is at a distance of 5 units to the left of the Y-axis, h = -5
∴ The equation of the required line is x = -5.
[Note: Answer given in the textbook is ‘y = -5
However, we found that ‘x = – 5’.]

iii. Equation of a line parallel to the X-axis is of the form y = k (k > 0 or k < 0).
Since the line is at a distance of 4 units from the point (- 2, 3),
k = 4 + 3 = 7 or k = 3- 4 = -1
∴ The equation of the required line is y = 1 or y = – 1.

Question 2.
Obtain the equation of the line:
i. parallel to the X-axis and making an intercept of 3 units on the Y-axis.
ii. parallel to the Y-axis and making an intercept of 4 units on the X-axis.
Solution:
i. Equation of a line parallel to X-axis with y-intercept ‘k’ isy = k.
Here, y-intercept = 3
∴ The equation of the required line is y = 3.

ii. Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.
Here, x-intercept = 4
∴ The equation of the required line is x = 4.

Question 3.
Obtain the equation of the line containing the point:
i. A(2, – 3) and parallel to the Y-axis.
ii. B(4, – 3) and parallel to the X-axis.
Solution:
i. Equation of a line parallel to Y-axis is of the form x = h.
Since the line passes through A(2, – 3), h = 2
∴ The equation of the required line is x = 2.

ii. Equation of a line parallel to X-axis is of the formy = k.
Since the line passes through B(4, – 3), k = -3
∴ The equation of the required line is y = – 3.

Question 4.
Find the equation of the line:
i. passing through the points A(2, 0) and B(3,4)
ii. passing through the points P(2, 1) and Q(2,-1)
Solution:
i. The required line passes through the points A(2, 0) and B(3,4).
Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
Here, (x₁y₁) = (2,0) and (x₁,y₂) = (3,4)
∴ The equation of the required line is
∴ \(\frac{y-0}{4-0}=\frac{x-2}{3-2}\)
∴ \(\frac{y}{4}=\frac{x-2}{1}\)
∴ y = 4(x – 2)
∴ y = 4x – 8
∴ 4x – y – 8 = 0

ii. The required line passes through the points P(2, 1) and Q(2,-1).
Since both the given points have same
x co-ordinates i.e. 2,
the given points lie on the line x = 2.
∴ The equation of the required line is x = 2.

Question 5.
Find the equation of the line:
i. containing the origin and having inclination 60°.
ii. passing through the origin and parallel to AB, where A is (2,4) and B is (1,7).
iii. having slope 1/2 and containing the point (3, -2)
iv. containing the point A(3, 5) and having slope 2/3
v. containing the point A(4, 3) and having inclination 120°.
vi. passing through the origin and which bisects the portion of the line 3JC + y = 6 intercepted between the co-ordinate axes.
Solution:
i. Given, Inclination of line = θ = 60°
Slope of the line (m) = tan θ = tan 60°
= \(\sqrt{3}\)
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
.‘. The equation of the required line is y = \(\sqrt{3}\) x

ii. Given, A (2, 4) and B (1, 7)
Slope of AB = \(\frac{7-4}{1-2}\) = -3 1-2
Since the required line is parallel to line AB, slope of required line (m) = slope of AB
∴ m = – 3 and the required line passes through the origin.
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is y = – 3x

iii. Given, slope(m) = \(=\frac{1}{2}\) and the line passes through (3, – 2).
Equation of the line in slope point form is
y-y ₁= m(x-x₁)
∴ The equation of the required line is
[y-(- 2)]=\(\frac{1}{2}\)(x-3)
∴ 2(y + 2)=x – 3
∴ 2y + 4 = x – 3
∴ x – 2y – 7 = 0

iv. Given, slope(m) = \(\frac{2}{3}\) and the line passes through (3, 5).
Equation of the line in slope point form is y-y₁ = m(x -x₁)
∴ The equation of the required line is y – 5 = \(\frac{2}{3}\)(x-3)
∴ 3 (y – 5) = 2 (x – 3)
∴ 3y – 15 = 2x – 6
∴ 2x – 3y + 9 = 0

v. Given, Inclination of line = θ = 120°
Slope of the line (m) = tan θ = tan 120°
= tan (90° + 30°)
= – cot 30°
= – \(\sqrt{3}\)
and the line passes through A(4, 3).
Equation of the line in slope point form is y-y₁ = m(x -x₁)
∴ The equation of the required line is
y- 3 = –\(\sqrt{3}\)(x-4)
∴ y – 3 = –\(\sqrt{3}\) x + 4\(\sqrt{3}\)
∴ \(\sqrt{3}\)x + y – 3 -4\(\sqrt{3}\) = 0

vi.

Given equation of the line is 3x +y = 6.
∴ \(\frac{x}{2}+\frac{y}{6}=1\)
This equation is of the form \(\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}\) = 1,
where a = 2, b = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2, 0) and B(0, 6) respectively. Required line is passing through the midpoint of AB.
∴ Midpoint of AB = ( \(\frac{2+0}{2}, \frac{0+6}{2}\) ) = (1,3)
∴ Required line passes through (0, 0) and (1,3).
Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-0}{3-0}=\frac{x-0}{1-0}\)
\(\frac{y}{3}=\frac{x}{1}\)
∴ y = 3x
∴ 3x – y = 0

Alternate Method:
Given equation of the line is 3x + y = 6 …(i)
Substitute y = 0 in (i) to get a point on X-axis.
∴ 3x + 0 = 6
∴ x = 2
Substitute x = 0 in (i) to get a point on Y-axis.
∴ 3(0) + 7 = 6
∴ y = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2,0) and B(0,6) respectively.
Let M be the midpoint of AB.
M = \(\left(\frac{2+0}{2}, \frac{0+6}{2}\right)\) = (1,3)
Slope of OM (m) = \(\frac{3-0}{1-0}\) = 3
Equation of OM is of the formy = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 6.
Line y = mx + c passes through the points A(2,1) and B(3,2). Determine m and c.
Solution:
Given, A(2, 1) and B(3,2)
Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-1}{2-1}=\frac{x-2}{3-2}\)
∴ \(\frac{y-1}{1}=\frac{x-2}{1}\)
∴ y – 1 = x – 2
∴ y = x – 1
Comparing this equation with y = mx + c, we get
m = 1 and c = – 1

Alternate Method:
Points A(2, 1) and B(3, 2) lie on the line y = mx + c.
∴ They must satisfy the equation.
∴ 2m + c = 1 …(i)
and 3m + c = 2 …(ii)
equation (ii) – equation (i) gives m = 1
Substituting m = 1 in (i), we get 2(1) + c = 1
∴ c = 1 – 2 = – 1

Question 7.
Find the equation of the line having inclination 135° and making x-intercept 7.
Solution:
Given, Inclination of line = 0 = 135°
∴ Slope of the line (m) = tan 0 = tan 135°
= tan (90° + 45°)
= – cot 45° = – 1 x-intercept of the required line is 7.
∴ The line passes through (7, 0).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
∴ The equation of the required line is y — 0 = – 1 (x – 7)
∴ y = -x + 7
∴ x + y – 7 = 0

Question 8.
The vertices of a triangle are A(3, 4), B(2, 0) and C(- 1, 6). Find the equations of the lines containing
i. side BC
ii. the median AD
iii. the midpoints of sides AB and BC.
Solution:
Vertices of AABC are A(3, 4), B(2, 0) and C(- 1, 6).
i. Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the side BC is
\(\frac{y-0}{6-0}=\frac{x-2}{-1-2}\)
\(\frac{y}{6}=\frac{x-2}{-3}\)
∴ – 3y = 6x – 12
∴ 6x + 3y – 12 = 0
∴ 2x + y – 4 = 0

ii. Let D be the midpoint of side BC.
Then, AD is the median through A.
∴ D = \(\left(\frac{2-1}{2}, \frac{0+6}{2}\right)=\left(\frac{1}{2}, 3\right)\)
The median AD passes through the points
A(3,4) and D( \(\frac{1}{2}\) , 3)

∴ The equation of the median AD is
\(\frac{y-4}{3-4}=\frac{x-3}{\frac{1}{2}-3}\)
\(\frac{y-4}{-1}=\frac{x-3}{-\frac{5}{2}}\)
\(\frac{5}{2}\)(y-4) = x – 3
∴ 5y – 20 = 2x – 6
∴ 2x – 5y + 14 = 0

iii. Let D and E be the midpoints of side AB and side BC respectively.
The equation of the line DE is

∴ -4(y-2) = 2x-5
∴ 2x + 4y – 13 = 0

Question 9.
Find the x and y-intercepts of the following lines:
i. \(\frac{x}{3}+\frac{y}{2}=1\)
ii. \(\frac{3 x}{2}+\frac{2 y}{3}=1\)
iii. 2x – 3y + 12 = 0
Solution:
i. Given equation of the line is latex]\frac{x}{3}+\frac{y}{2}=1[/latex]
This is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = 3, y-intercept = 2

ii. Given equation of the line is \(\frac{3 x}{2}+\frac{2 y}{3}\) = 1
∴ \(\frac{x}{\left(\frac{2}{3}\right)}+\frac{y}{\left(\frac{3}{2}\right)}\) = 1
This is of the form = \(\frac{x}{a}+\frac{y}{b}\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = \(\frac{2}{3}\) and y-intercept = \(\frac{3}{2}\)

iii. Given equation of the line is 2x – 3y + 12 = 0
∴ 2x – 3y = – 12
∴ \(\frac{2 x}{(-12)}-\frac{3 y}{(-12)}=1\)
∴ \(\frac{x}{-6}+\frac{y}{4}=1\)
This is of the form \(\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = – 6 and y-intercept = 4

Question 10.
Find equations of the line which contains the point A(l, 3) and the sum of whose intercepts on the co-ordinate axes is zero.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be
\(\frac{x}{a}+\frac{y}{b}=1\) ………..(i)
Since, the sum of the intercepts of the line is zero.
∴ a + b = 0
∴ b = – a
Substituting b = – a in (i), we get
\(\frac{x}{a}+\frac{y}{(-a)}=1\)
x – y = a .. .(ii)
Since, the line passes through A(1, 3).
∴ 1 – 3 = a
∴ a = – 2
Substituting the value of a in (ii), equation of the required line is
∴ x – y = – 2,
∴ x – y + 2 = 0

Case II: Line passing through origin.
Slope of line passing through origin and
A(1, 3) is m = \(\frac{3-0}{1-0}\) = 3
∴ Equation of the line having slope m and passing through origin (0, 0) is / = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 11.
Find equations of the line containing the point A(3, 4) and making equal intercepts on the co-ordinate axes.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\) …………(i)
This line passes through A(3, 4).
∴ \(\frac{3}{a}+\frac{4}{b}=1\)……………..(ii)
Since, the required line make equal intercepts on the co-ordinate axes.
∴ a = b …(iii)
Substituting the value of b in (ii), we get
\(\frac{3}{a}+\frac{4}{a}=1\)
∴ \(\frac{7}{a}=1\)
∴ a = 7
∴ b = 7 …[From (iii)]
Substituting the values of a and b in (i), equation of the required line is
\(\frac{x}{7}+\frac{y}{7}=1\) = 1
∴ x + y = 7

Case II: Line passing through origin.
Slope of line passing through origin and A(3,4) is m = \(=\frac{4-0}{3-0}=\frac{4}{3}\)
∴ Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is 4
y = \(\frac{4}{3}\)x
∴ 4x – 3y = 0

Question 12.
Find the equations of the altitudes of the triangle whose vertices are A(2, 5), B(6, – 1 ) and C(- 4, – 3).
Solution:

A(2, 5), B(6, – 1), C(- 4, – 3) are the vertices of ∆ABC.
Let AD, BE and CF be the altitudes through the vertices A, B and C respectively of ∆ABC.
∴ Slope of AD = -5 …[∵AD ⊥ BC]
Since altitude AD passes through (2, 5) and has slope – 5,
equation of the altitude AD is y – 5 = -5 (x – 2)
∴ y – 5 = – 5x + 10
∴ 5x +y -15 = 0
Now, slope of AC = \(\frac{-3-5}{-4-2}=\frac{-8}{-6}=\frac{4}{3}\)
Slope of BE = \(\frac{-3}{4}\)
…[∵ BE ⊥ AC]
Since altitude BE passes through (6,-1) and has slope \(\frac{-3}{4}\),
equation of the altitude BE is
y-(-1) = \(\frac{-3}{4}\) (x – 6)
∴ 4 (y + 1) = – 3 (x – 6)
∴ 4y + 4 =-3x+ 18
∴ 3x + 4y – 14 = 0
Also, slope of AB = \(\frac{-1-5}{6-2}=\frac{-6}{4}=\frac{-3}{2}\)
∴ Slope of CF = \({2}{3}\) ….[∵ CF ⊥ AB]
Since altitude CF passes through (- 4, – 3) and has slope , \(\frac{2}{3}\)
equation of the altitude CF is
y-(-3) = \(\frac{2}{3}\)[x-(-4)]
∴ 3 (y + 3) = 2 (x + 4)
∴ 3y + 9 = 2x + 8
∴ 2x – 3y – 1 = 0

Question 13.
Find the equations of perpendicular bisectors of sides of the triangle whose vertices are P(-1, 8), Q(4, – 2) and R(- 5, – 3).
Solution:

Let A, B and C be the midpoints of sides PQ, QR and PR respectively of APQR.
A is the midpoint of side PQ.

Slope of perpendicular bisector of PQ is \(\frac{1}{2}\) and it passes through (\(\frac{3}{2}\)), 3).
Equation of the perpendicular bisector of side PQ is
y – 3 = \(\frac{1}{2}\)(x – \(\frac{3}{2}\))
y – 3 = (\(\frac{1}{2}\left(\frac{2 x-3}{2}\right)\))
∴ 4(y – 3) = 2x – 3
∴ 4y – 12 = 2x – 3
∴ 2x – 4y + 9 = 0
B is the midpoint of side QR
∴ B = \(\left(\frac{4-5}{2}, \frac{-2-3}{2}\right)=\left(\frac{-1}{2}, \frac{-5}{2}\right)\)
Slope of side QR = \(\frac{-3-(-2)}{-5-4}=\frac{-1}{-9}=\frac{1}{9}\)
∴ Slope of perpendicular bisector of QR is -9 and it passes through \(\left(-\frac{1}{2},-\frac{5}{2}\right)\)
∴ Equation of the perpendicular bisector of side QR is

∴ 2y + 5 = -18x – 9
∴ 18x + 2y + 14 = 0
∴ 9x + y + 7 = 0
C is the midpoint of side PR.

Equation of the perpendicular bisector of PR is \(y-\frac{5}{2}=-\frac{4}{11}(x+3)\)
∴ \(11\left(\frac{2 y-5}{2}\right)\) =-4(x + 3)
∴ 11(2y – 5) = – 8 (x + 3)
∴ 22y – 55 = – 8x – 24
∴ 8x + 22y -31 = 0

Question 14.
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(2, – 2), B(l, 1) and C(-1,0).
Solution:
Let O be the orthocentre of AABC.
Let AM and BN be the altitudes of sides BC and AC respectively.
Now, slope of BC = \(\frac{0-1}{-1-1}=\frac{-1}{-2}=\frac{1}{2}\)
Slope of AM = -2 ,..[∵ AM ⊥ BC]
Since AM passes through (2, – 2) and has slope -2,
equation of the altitude AM is y – (- 2) = – 2 (x – 2)
∴ y + 2 = -2x + 4
∴ 2x + y – 2 = 0 …(i)

Also, slope of AC = \(\frac{0-(-2)}{-1-2}=\frac{2}{-3}\)
∴ Slope of BN = \(\frac{3}{2}\) …[∵ BN ⊥ AC]
Since BN passes through (1,1) and has slope \(\frac{3}{2}\), equation of the altitude BN is
y – 1 = \(\frac{3}{2}\)(x-1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0 …(ii)
To find co-ordinates of orthocentre, we have to solve equations (i) and (ii).
By (i) x 2 + (ii), we get
7x – 5 = 0
∴ x = \(\frac{5}{7}\)
substituting x = \(\frac{5}{7}\) in eq (i), we get
2(\(\frac{5}{7}\)) + y – 2 = 0
∴ y = -2(\(\frac{5}{7}\)) + 2
∴ y = \(\frac{-10+14}{7}=\frac{4}{7}\)
∴ Coordinates of orthocentre O = \(\left(\frac{5}{7}, \frac{4}{7}\right)\)

Question 15.
N(3, – 4) is the foot of the perpendicular drawn from the origin to line L. Find the equation of line L.
Solution:

Slope of ON = \(\frac{-4-0}{3-0}=\frac{-4}{3}\)
Since line L ⊥ ON,
slope of the line L is \(\frac{3}{4}\) and it passes through point N(3, -4).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
Equation of line L is
y-(-4) = \(\frac{3}{4}\)(x-3)
∴ 4(y + 4) = 3(x – 3)
∴ 4y + 16 = 3x – 9
∴ 3x – 4y – 25 = 0

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.2

Question 1.
Find the slope of each of the lines which passes through the following points:
i. A(2, -1), B(4,3)
ii. C(- 2,3), D(5, 7)
iii. E(2,3), F(2, – 1)
iv. G(7,1), H(- 3,1)
Solution:
i. Here, A = (2, -1) andB = (4, 3)
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-(-1)}{4-2}=\frac{4}{2}\) = 2

ii. Here, C = (-2, 3) and D = (5, 7)
Slope of line CD = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{7-3}{5-(-2)}=\frac{4}{7}\)

iii. Here, E s (2, 3) and F = (2, -1)
Slope of line EF = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-1-3}{2-2}=\frac{-4}{0}\), which ix not defined.

Alternate Method:
Points E and F have same x co-ordinates i.e. 2.
Points E and F lie on a line parallel to Y-axis.
∴ The slope of EF is not defined.

iv. Here, G = (7, 1) and H = (-3, 1)
Slope of line GH = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1-1}{-3-7}\) = o

Alternate Method:
Points G and H have same y co-ordinate i.e. 1.
∴ Points G and H lie on a line parallel to the
X-axis.
∴ The slope of GH is 0.

Question 2.
If the x and x-intercepts of line L are 2 and 3 respectively, then find the slope of line L.
Solution:
Given, x-intercept of line L is 2 and y-intercept of line L is 3
∴ The line L intersects X-axis at (2, 0) and Y-axis at (0,3).
∴ The line L passes through (2, 0) and (0, 3).
Slope of line L = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{0-2}=\frac{-3}{2}\)

Question 3.
Find the slope of the line whose inclination is 30°.
Solution:
Given, inclination (θ) = 30°
∴ Slope of the line = tanθ = tan30° = \(\frac{1}{\sqrt{3}}\)

Question 4.
Find the slope of the line whose inclination is \(\frac{\pi}{4}\)
Solution:
Given, inclination (0) = \(\frac{\pi}{4}\)
∴ Slope of the line = tan θ = tan\(\frac{\pi}{4}\) = 1

Question 5.
A line makes intercepts 3 and 3 on the co-ordinate axes. Find the inclination of the line.
Solution:
Given, x-intercept of line is 3 and y-intercept of line is 3
∴ The line intersects X-axis at (3, 0) and Y-axis at (0, 3).
∴ The line passes through (3, 0) and (0,3).
∴ Slope of line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{0-3}\) = -1
But, slope of a line = tan θ
∴ tan θ = – 1
= – tan \(\frac{\pi}{4}\)
= tan(π-\(\frac{\pi}{4}\) ) …[v tan(π – θ) = -tan θ]
tan θ = tan \(\frac{3\pi}{4}\)
θ = \(\frac{3\pi}{4}\)
The inclination of the line is \(\frac{3\pi}{4}\).
[Note: Answer given in the textbook is ‘-1 However, as per our calculation it is \(\frac{3\pi}{4}\)]

Question 6.
Without using Pythagoras theorem, show that points A (4, 4), B (3, 5) and C (- 1, – 1) are the vertices of a right-angled triangle.
Solution:
Given, A(4,4), B(3, 5), C (-1, -1).
Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-4}{3-4}\) = – 1
Slope of BC = \(\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}\)
Slope of AC = \(\frac{-1-4}{-1-4}\) = 1
Slope of AB x slope of AC = – 1 x 1 = – 1
∴ side AB ⊥ side AC
∴ ∆ABC is a right angled triangle right angled at A.
∴ The given points are the vertices of a right angled triangle.

Question 7.
Find the slope of the line which makes angle of 45° with the positive direction of the Y-axis measured anticlockwise.
Solution:

Since the line makes an angle of 45° with positive direction of Y-axis in anticlockwise direction,
Inclination of the line (0) = (90° + 45°)
∴ Slope of the line = tan(90° + 45°)
= – cot 45°
= -1

Question 8.
Find the value of k for which the points P(k, -1), Q(2,1) and R(4,5) are collinear.
Solution:
Given, points P(k, – 1), Q (2, 1) and R(4, 5) are collinear.
∴ Slope of PQ = Slope of QR .
∴ \(\frac{1-(-1)}{2-k}=\frac{5-1}{4-2}\)
∴ \(\frac{2}{2-k}=\frac{4}{2}\)
∴ 4 = 4 (2 – k)
∴ 1 = 2 – k
∴ k = 2 – 1 = 1

Question 9.
Find the acute angle between the X-axis and the line joining the points A(3, -1) and B(4, – 2).
Solution:
Given, A (3, – 1) and B (4, – 2)
Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-(-1)}{4-3}\) = – 1
But, slope of a line = tan θ
∴ tan θ = – 1
= – tan 45°
= tan (180° -45°)
… [∵ tan (180° – θ) = -tan θ]
= tan 135°
∴ θ = 135°

Let α be the acute angle that line AB makes with X-axis.
Then, α + 0 = 180°
α = 180°- 135° = 45°
∴ The acute angle between the X-axis and the line joining the points A and B is 45°.

Question 10.
A line passes through points A(xi, y0 and B(h, k). If the slope of the line is m, then show that k – y₁ = m (h – x₁).
Solution:
Given, A(x₁, y₁), B(h, k) and
slope of line AB = m
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
∴ m = \(\frac{\mathrm{k}-y_{1}}{\mathrm{~h}-x_{1}}\)
∴ k – y₁ = m (h – x₁)

Question 11.
If the points A(h, 0), B(0, k) and C(a, b) lie on a line, then show that \(\frac{a}{h}+\frac{b}{k}\) = 1. ‘
Solution:
Given, A(h, 0), B(0, k) and C(a, b)
Since the points A, B and C lie on a line, they are collinear.
∴ Slope of AB = slope of BC

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.1

Question 1.
If A(1, 3) and B(2, 1) are points, find the equation of the locus of point P such that PA = PB.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(1, 3), B(2, 1) and
PA = PB
∴ PA² = PB²
∴ (x – 1)² + ( y – 3)² = (x – 2)² + (y – 1)²
∴ x² – 2x + 1 + y² – 6y + 9 = x² – 4x + 4 + y² – 2y + 1
-2x – 6y + 10 = -4x – 2y + 5
∴ 2x – 4y + 5 = 0
∴ The required equation of locus is 2x – 4y + 5 = 0.

Question 2.
A(- 5,2) and B(4,1). Find the equation of the locus of point P, which is equidistant from A and B.
Solution:
Let P(x, y) be any point on the required locus.
P is equidistant from A(- 5, 2) and B(4, 1).
∴ PA = PB
∴ PA² = PB²
∴ (x + 5)² + (y – 2)² = (x – 4)² + (y – 1)²
∴ x² + 10x + 25 + y² — 4y + 4
= x² – 8x + 16 + y² – 2y + 1
∴ 10x – 4y + 29 = -8x – 2y + 17
∴ 18x – 2y + 12 = 0
∴ 9x – y + 6 = 0
The required equation of locus is 9x -y + 6 = 0.

Question 3.
If A(2, 0) and B(0, 3) are two points, find the equation of the locus of point P such that AP = 2BP.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(2, 0), B(0, 3) and AP = 2BP
∴ AP² = 4BP²
∴ (x – 2)² + (y – 0)² = 4[(x – 0)² + (y – 3)²]
∴ x² – 4x + 4 + y² = 4(x² + y² – 6y + 9)
x² – 4x + 4 + y² = 4x² + 4y² – 24y + 36
∴ 3x² + 3 y² + 4x – 24y + 32 = 0
∴ The required equation of locus is
3x² + 3y² + 4x – 24y + 32 = 0.
[Note: Answer given in the textbook , is
‘3x² + 3y² + 4x + 24y + 32 = O’.
However, as per our calculation it is ‘3x² + 3y² + 4x – 24y + 32 = 0’.]

Question 4.
If A(4,1) and B(5,4), find the equation of the locus of point P such that PA² = 3PB².
Solution:
Let P(x, y) be any point on the required locus. Given, A(4,1), B(5,4) and PA2 = 3PB2
∴ (x – 4)2 + (y – 1)2 = 3[(x – 5)² + (y – 4)²]
∴ x² – 8x + 16 + y² – 2y + 1 = 3(x² – 10x + 25 + y² – 8y + 16)
∴ x² – 8x + y² – 2y + 17 = 3x² -30x + 75 + 3y² – 24y + 48
∴ 2x² + 2y² – 22x – 22y + 106 = 0
∴ x² + y² – 11x – 11y + 53 = 0
∴ The required equation of locus is
x² + y² – 11x – 11y + 53 = 0.

Question 5.
A(2, 4) and B(5, 8), find the equation of the
locus of point P such that PA² – PB² = 13.
Solution:
Let P(x, y) be any point on the required locus. Given, A(2,4), B(5, 8) and PA²-PB² = 13
∴ [(x -2)² + (y – 4)²] – [(x -5)² + (y- 8)²] = 13
∴ (x² – 4x + 4 + y² – 8y + 16) – (x² – 10x + 25 + y² – 16y + 64) =13
∴ x² – 4x+ y² – 8y + 20 – x² + 10x – y² + 16y – 89 = 13
∴ 6x + 8y- 69 = 13
∴ 6x + 8y – 82 = 0
∴ 3x + 4y – 41 = 0
∴ The required equation of locus is 3x + 4y- 41 = 0.

Question 6.
A(1, 6) and B(3, 5), find the equation of the locus of point P such that segment AB subtends right angle at P. (∠APB = 90°)
Solution:
Let P(x, y) be any point on the required locus. Given,
A(l, 6) and B(3, 5),
∠APB = 90°
∴ ΔAPB is a right angled triangle,
By Pythagoras theorem,

AP² + PB² = AB² P (x,y)
∴ [(x – 1)² + (y – 6)²] + [(x – 3)² + (y – 5)²] = (1 – 3)² + (6 -5)²
∴ x² — 2x + 1 + y² — 12y + 36 + x² – 6x + 9 + y² – 10y + 25 = 4 + 1
∴ 2x² + 2y² – 8x – 22y + 66 = 0
∴ x² + y² – 4x – 11y + 33 = 0
∴ The required equation of locus is x² + y² – 4x – 11y + 33 = 0.
[Note: Answer given in the textbook is
‘3x² + 4y² – 4x – 11y + 33 = 0’.
However, as per our calculation it is ‘x² + y² – 4x – 11y + 33 = O

Question 7.
If the origin is shifted to the point 0′(2, 3), the axes remaining parallel to the original axes, find the new co-ordinates of the points
i. A(1, 3) ii. B(2,5)
Solution:
Origin is shifted to (2, 3) = (h, k)
Let the new co-ordinates be (X, Y).
x = X + handy = Y + k
x = X + 2 andy = Y + 3 …(i)

i. Given, A(x, y) = A( 1, 3)
x = X + 2 andy = Y + 3 …[From(i)]
∴ 1 = X + 2 and 3 = Y + 3 X = – 1 and Y = 0
∴ The new co-ordinates of point A are (- 1,0).

ii. Given, B(x, y) = B(2, 5)
x = X + 2 and y = Y + 3 …[From(i)]
∴ 2 = X + 2 and 5 = Y + 3
∴ X = 0 and Y = 2
∴ The new co-ordinates of point B are (0, 2).

Question 8.
If the origin is shifted to the point O'(1, 3), the axes remaining parallel to the original axes, find the old co-ordinates of the points
i. C(5,4) ii. D(3,3)
Solution:
Origin is shifted to (1,3) = (h, k)
Let the new co-ordinates be (X, Y).
x = X + h andy = Y + k
∴ x = X+1 andy = Y + 3 …(i)

i. Given, C(X, Y) = C(5, 4)
x = X +1 andy = Y + 3 …[From(i)]
∴ x = 5 + 1 = 6 andy = 4 + 3 = 7
∴ The old co-ordinates of point C are (6, 7).

ii. Given, D(X, Y) = D(3, 3)
x = X + 1 andy = Y + 3 …[From(i)]
∴ x = 3 + 1 = 4 and y = 3 + 3 = 6
∴ The old co-ordinates of point D are (4, 6).

Question 9.
If the co-ordinates A(5, 14) change to B(8, 3) by shift of origin, find the co-ordinates of the point, where the origin is shifted.
Solution:
Let the origin be shifted to (h, k).
Given, A(x, y) = A(5,14), B(X, Y) = B(8, 3)
Since x = X + h andy = Y + k,
5 = 8 + hand 14 = 3 + k ,
∴ h = – 3 and k = 11
The co-ordinates of the point, where the origin is shifted are (- 3, 11).

Question 10.
Obtain the new equations of the following loci if the origin is shifted to the point 0′(2,2), the direction of axes remaining the same:
i. 3x-y + 2 = 0
11. x²+y²-3x = 7
iii. xy – 2x – 2y + 4 = 0
iv. y² – 4x – 4y + 12 = 0
Solution:
Given, (h,k) = (2,2)
Let (X, Y) be the new co-ordinates of the point (x,y).
∴ x = X + handy = Y + k
∴ x = X + 2 andy = Y + 2
i. Substituting the values of x and y in the equation 3x -y + 2 = 0, we get
3(X + 2) – (Y + 2) + 2 = 0
∴ 3X + 6-Y-2 + 2 = 0
∴ 3 X – Y + 6 = 0, which is the new equation of locus.

ii. Substituting the values of x and y in the equation
x² + y² – 3x = 7, we get
(X + 2)² + (Y + 2)² – 3(X + 2) = 7
∴ X² + 4X + 4 + Y² + 4Y + 4 – 3X – 6 = 7
∴ X² + Y² + X + 4Y – 5 = 0, which is the new
equation of locus.

iii; Substituting the values of x and y in the equation xy – 2x – 2y + 4 = 0, we get
(X + 2) (Y + 2) – 2(X + 2) – 2(Y + 2) + 4 = 0
∴ XY + 2X + 2Y + 4 – 2X – 4-2Y- 4 + 4 = 0
∴ XY = 0, which is the new equation of locus.

iv. Substituting the values of x and y in the equation y² – 4x – 4y + 12 = 0, we get
(Y + 2)² – 4(X + 2) – 4(Y + 2) + 12 = 0
∴ Y² + 4Y + 4 – 4X – 8 – 4Y -8 + 12 = 0
∴ Y² – 4X = 0, which is the new equation of locus.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

(I) Select the correct option from the given alternatives.

Question 1.
Given A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 2
\end{array}\right]\), I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) if A – λI is a singular matrix, then ___________
(a) λ = 0
(b) λ² – 3λ – 4 = 0
(c) λ² + 3λ – 4 = 0
(d) λ² – 3λ – 6 = 0
Answer:
(b) λ² – 3λ – 4 = 0
Hint:

Question 2.
Consider the matrices A = \(\left[\begin{array}{ccc}
4 & 6 & -1 \\
3 & 0 & 2 \\
1 & -2 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 4 \\
0 & 1 \\
-1 & 2
\end{array}\right]\), C = \(\left[\begin{array}{l}
3 \\
1 \\
2
\end{array}\right]\). Out of the given matrix products, ___________
(i) (AB)^TC
(ii) C^TC(AB)^T
(iii) C^TAB
(iv) A^TABB^TC
(a) Exactly one is defined
(b) Exactly two are defined
(c) Exactly three are defined
(d) all four are defined
Answer:
(c) Exactly three are defined
Hint:
A is of order 3 × 3, B is of order 3 × 2 and C is of order 3 × 1.
(AB)^TC is of order 2 × 1.
C^TC and (AB)^T are of different orders.
C^TC (AB)^T is not defined.
C^TAB is of order 1 × 2.
A^TABB^TC is of order 3 × 1.

Question 3.
If A and B are square matrices of equal order, then which one is correct among the following?
(a) A + B = B + A
(b) A + B = A – B
(c) A – B = B – A
(d) AB = BA
Answer:
(a) A + B = B + A
Hint:
Matrix addition is commutative.
∴ A + B = B + A

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
a & 2 & b
\end{array}\right]\) is a matrix satisfying the equation AA^T = 9I, where I is the identity matrix of order 3, then the ordered pair (a, b) is equal to ___________
(a) (2, -1)
(b) (-2, 1)
(c) (2, 1)
(d) (-2, -1)
Answer:
(d) (-2, -1)

Question 5.
If A = \(\left[\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right]\) and |A³| = 125, then α = ___________
(a) ±3
(b) ±2
(c) ±5
(d) 0
Answer:
(a) ±3
Hint:
|A³| = 125
|A|³ = 53 …….[∵ |Aⁿ| = |A|ⁿ, n ∈ N]
∴ |A| = 5
\(\left|\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right|=5\)
α² – 4 = 5
α² = 9
∴ α = ± 3

Question 6.
If \(\left[\begin{array}{ll}
5 & 7 \\
x & 1 \\
2 & 6
\end{array}\right]-\left[\begin{array}{cc}
1 & 2 \\
-3 & 5 \\
2 & y
\end{array}\right]=\left[\begin{array}{cc}
4 & 5 \\
4 & -4 \\
0 & 4
\end{array}\right]\), then ___________
(a) x = 1, y = -2
(b) x = -1, y = 2
(c) x = 1, y = 2
(d) x = -1, y = -2
Answer:
(c) x = 1, y = 2

Question 7.
If A + B = \(\left[\begin{array}{ll}
7 & 4 \\
8 & 9
\end{array}\right]\) and A – B = \(\left[\begin{array}{ll}
1 & 2 \\
0 & 3
\end{array}\right]\), then the value of A is ___________
(a) \(\left[\begin{array}{ll}
3 & 1 \\
4 & 3
\end{array}\right]\)
(b) \(\left[\begin{array}{ll}
4 & 3 \\
4 & 6
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
6 & 2 \\
8 & 6
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
7 & 6 \\
8 & 12
\end{array}\right]\)
Answer:
(b) \(\left[\begin{array}{ll}
4 & 3 \\
4 & 6
\end{array}\right]\)

Question 8.
If \(\left[\begin{array}{cc}
x & 3 x-y \\
z x+z & 3 y-w
\end{array}\right]=\left[\begin{array}{ll}
3 & 2 \\
4 & 7
\end{array}\right]\), then ___________
(a) x = 3, y = 7, z = 1, w = 14
(b) x = 3, y = -5, z = -1, w = -4
(c) x = 3, y = 6, z = 2, w = 7
(d) x = -3, y = -7, z = -1, w = -14
Answer:
(a) x = 3, y = 7, z = 1, w = 14

Question 9.
For suitable matrices A, B, the false statement is ___________
(a) (AB)^T = A^TB^T
(B) (A^T)^T = A
(C) (A – B)^T = A^T – B^T
(D) (A + B)^T = A^T + B^T
Answer:
(a) (AB)^T = A^TB^T
Hint:
(AB)^T = B^TA^T

Question 10.
If A = \(\left[\begin{array}{cc}
-2 & 1 \\
0 & 3
\end{array}\right]\) and f(x) = 2x² – 3x, then f(A) = ___________
(a) \(\left[\begin{array}{cc}
14 & 1 \\
0 & -9
\end{array}\right]\)
(b) \(\left[\begin{array}{cc}
-14 & 1 \\
0 & 9
\end{array}\right]\)
(c) \(\left[\begin{array}{cc}
14 & -1 \\
0 & 9
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
-14 & -1 \\
0 & -9
\end{array}\right]\)
Answer:
(c) \(\left[\begin{array}{cc}
14 & -1 \\
0 & 9
\end{array}\right]\)
Hint:

(II) Answer the following questions.

Question 1.
If A = diag[2, -3, -5], B = diag[4, -6, -3] and C = diag [-3, 4, 1], then find
i. B + C – A
ii. 2A + B – 5C.
Solution:

Question 2.
If f(α) = A = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\), find
i. f(-α)
ii. f(-α) + f(α)
Solution:

Question 3.
Find matrices A and B, where
(i) 2A – B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\) and A + 3B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\)
(ii) 3A – B = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 0 & 5
\end{array}\right]\) and A + 5B = \(\left[\begin{array}{ccc}
0 & 0 & 1 \\
-1 & 0 & 0
\end{array}\right]\)
Solution:
i. Given equations are
2A – B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\) …….(i)

Question 4.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
3 & -2 \\
-1 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & 4 & 1 \\
2 & -1 & -3
\end{array}\right]\), verify
i. (A + 2B^T)^T = A^T + 2B
ii. (3A – 5B^T)^T = 3A^T – 5B.
Solution:

Question 5.
If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) and A + A^T = I, where I is a unit matrix of order 2 × 2, then find the value of α.
Solution:

Question 6.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & 2 \\
-1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 3 & 2 \\
4 & -1 & -3
\end{array}\right]\), show that AB is singular.
Solution:

Question 7.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 4 & 6 \\
1 & 2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
-3 & 2 & -1 \\
-2 & 1 & 0
\end{array}\right]\), show that AB and BA are both singular matrices.
Solution:

Question 8.
If A = \(\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right]\), show that BA = 6I.
Solution:

Question 9.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
0 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -2
\end{array}\right]\), verify that |AB| = |A|.|B|.
Solution:

Question 10.
If Aα = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that Aα . Aβ = Aα+β
Solution:

Question 11.
If A = \(\left[\begin{array}{cc}
1 & \omega \\
\omega^{2} & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
\omega^{2} & 1 \\
1 & \omega
\end{array}\right]\), where ω is a complex cube root of unity, then show that AB + BA + A – 2B is a null matrix.
Solution:
ω is the complex cube root of unity.
ω³ = 1
ω³ – 1 = 0
(ω – 1) (ω² + ω + 1) = 0
ω = 1 or ω² + ω + 1 = 0
But, ω is a complex number.
1 + ω + ω² = 0 …….(i)
AB + BA + A – 2B

which is a null matrix.

Question 12.
If A = \(\left[\begin{array}{lrr}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & -3
\end{array}\right]\), show that A² = A.
Solution:

Question 13.
If A = \(\left[\begin{array}{ccc}
4 & -1 & -4 \\
3 & 0 & -4 \\
3 & -1 & -3
\end{array}\right]\), show that A² = I.
Solution:

Question 14.
If A = \(\left[\begin{array}{cc}
3 & -5 \\
-4 & 2
\end{array}\right]\), show that A² – 5A – 14I = 0.
Solution:

Question 15.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
-1 & 2
\end{array}\right]\), show that A – 4A + 3I = 0.
Solution:

Question 16.
If A = \(\left[\begin{array}{cc}
-3 & 2 \\
2 & -4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & x \\
y & 0
\end{array}\right]\) and (A + B)(A – B) = A² – B², find x and y.
Solution:
(A + B)(A – B) = A² – B²
A² – AB + BA – B² = A² – B²
-AB + BA = 0
AB = BA

By equality of matrices, we get
2 – 4x = -3x
∴ x = 2 and 2y = 2x
y = x
∴ y = 2
∴ x = 2, y = 2

Question 17.
If A = \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\), show that (A + B)(A – B) ≠ A² – B².
Solution:
We have to prove that
(A + B) . (A – B) ≠ A² – B²
i.e., to prove that A(A – B) + B(A – B) ≠ A² – B²
i.e., to prove that A² – AB + BA – B² ≠ A² – B²
i.e., to prove that AB ≠ BA.

∴ AB ≠ BA

Question 18.
If A = \(\left[\begin{array}{ll}
2 & -1 \\
3 & -2
\end{array}\right]\), find A³.
Solution:

∴ A² = I
Multiplying throughout by A, we get
A³ = A . I
∴ A³ = A

Question 19.
Find x, y if,

Solution:

Question 20.
Find x, y, z if

Solution:

Question 21.
If A = \(\left[\begin{array}{ccc}
2 & 1 & -3 \\
0 & 2 & 6
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & 0 & -2 \\
3 & -1 & 4
\end{array}\right]\), find AB^T and A^TB.
Solution:

Question 22.
If A = \(\left[\begin{array}{cc}
2 & -4 \\
3 & -2 \\
0 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 1 & 0
\end{array}\right]\), show that (AB)^T = B^TA^T.
Solution:

Question 23.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\), for all n ∈ N.
Solution:

Question 24.
If A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{cc}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\), for all n ∈ N.
Solution:

Question 25.
Two farmers Shantaram and Kantaram cultivate three crops rice, wheat, and groundnut. The sale (in Rupees) of these crops by both the farmers for the month of April and may 2008 is given below,

Find
(i) the total sale in rupees for two months of each farmer for each crop.
(ii) the increase in sales from April to May for every crop of each farmer.
Solution:
(i) Total sale for Shantaram:
For rice = 15000 + 18000 = ₹ 33000.
For wheat = 13000 + 15000 = ₹ 28000.
For groundnut = 12000 + 12000 = ₹ 24000.
Total sale for Kantaram:
For rice = 18000 + 21000 = ₹ 39000
For wheat = 15000 + 16500 = ₹ 31500
For groundnut = 8000 + 16000 = ₹ 24000

Alternate method:
Matrix form

∴ The total sale of April and May of Shantaram in ₹ is ₹ 33000 (rice), ₹ 28000 (wheat), ₹ 24000 (groundnut), and that of Kantaram in ₹ is ₹ 39000(rice), ₹ 31500(wheat), and ₹ 24000 (groundnut).

(ii) Increase in sale from April to May for Shantaram:
For rice = 18000 – 15000 = ₹ 3000
For wheat = 15000 – 13000 = ₹ 2000
For groundnut = 12000 – 12000 = ₹ 0
Increase in sale from April to May for Kantaram:
For rice = 21000 – 18000 = ₹ 3000
For wheat = 16500 – 15000 = ₹ 1500
For groundnut = 16000 – 8000 = ₹ 8000

Alternate method:
Matrix form

∴ The increase in sales for Shantaram from April to May in each crop is ₹ 3000 (rice), ₹ 2000(wheat), 0 (groundnut), and that for Kantaram is ₹ 3000 (rice), ₹ 1500 (wheat), and ₹ 8000 (groundnut).

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.7 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.7

Question 1.
Find A^T, if
i. A = \(\left[\begin{array}{cc}
1 & 3 \\
-4 & 5
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
2 & -6 & 1 \\
-4 & 0 & 5
\end{array}\right]\)
Solution:
i. A = \(\left[\begin{array}{cc}
1 & 3 \\
-4 & 5
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{rr}
1 & -4 \\
3 & 5
\end{array}\right]\)

ii. A = \(\left[\begin{array}{ccc}
-4 & 0 & 5
2 & -6 & 1 \\
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{cc}
2 & -4 \\
-6 & 0 \\
1 & 5
\end{array}\right]\)

[Note: Answer given in the textbook is A^T = \(\left[\begin{array}{cc}
2 & -4 \\
6 & 0 \\
1 & 5
\end{array}\right]\). However, as per our calculation it is A^T = \(\left[\begin{array}{cc}
2 & -4 \\
-6 & 0 \\
1 & 5
\end{array}\right]\). ]

Question 2.
If [a_ij]_3×3 where a_ij = 2(i – j), find A and
A^T. State whether A and A^T are symmetric or skew-symmetric matrices?
Solution:
A = [a_ij]_3×3 = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
Given a_ij = 2 (i — j)
∴ a₁₁ = 2(1-1) = 0,
a₁₂ = 2(1-2) = -2,
a₁₃ = 2(1-3) = -4,
a₂₁ = 2(2-1) = 2,
a₂₂ = 2(2-2) = 0,
a₂₃=2(2-3) = -2,
a₃₁ = 2(3-1) = 4,
a₃₂ = 2(3-2) = 2,
a₃₃=2(3-3) = 0

∴ A^T = -A and A = -A^T
∴ A and A^T both are skew-symmetric matrices.

Questionn 3.
If A = \(\left[\begin{array}{cc}
5 & -3 \\
4 & -3 \\
-2 & 1
\end{array}\right]\), prove that (2A)^T = 2A^T.
Solution:

From (i) and (ii), we get
(2A)^T = 2A^T

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\), prove that (3A)^T = 3A^T.
Solution:

From (i) and (ii), we get
(3A)^T = 3A^T

Question 5.
If A = \(\left[\begin{array}{ccc}
0 & 1+2 i & 1-2 \\
-1-2 i & 0 & -7 \\
2-i & 7 & 0
\end{array}\right]\),
where i = \(\sqrt{-1}\), prove that A^T = – A.
Solution:

Question 6.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\) , B = \(\left[\begin{array}{cc}
2 & 1 \\
4 & -1 \\
-3 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 4 \\
-2 & 3
\end{array}\right]\) then show that
i. (A + B)^T = AT + B^T
ii. (A – C)^T = A^T – C^T
Solution:

From (i) and (ii), we get
(A + B)^T = AT + B^T
[Note: The question has been modified.]

From (i) and (ii), we get
(A – C)^T = A^T – C^T</sup

Question 7.
If A = \(\left[\begin{array}{cc}
5 & 4 \\
-2 & 3
\end{array}\right]\) and \(\left[\begin{array}{cc}
-1 & 3 \\
4 & -1
\end{array}\right]\) then find C^T, such that 3A – 2B + C = I, where I is the unit matrix of order 2.
Solution:
3A – 2B + C = I
∴ C = I + 2B – 3A

Question 8.
If A = \(\left[\begin{array}{ccc}
7 & 3 & 0 \\
0 & 4 & -2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
0 & -2 & 3 \\
2 & 1 & -4
\end{array}\right]\), then find
i. A^T + 4B^T
ii. 5A^T – 5B^T
Solution:

ii. ii. 5A^T – 5B^T = 5(A^T – B^T)

Question 9.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
3 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
2 & 1 & -4 \\
3 & 5 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
0 & 2 & 3 \\
-1 & -1 & 0
\end{array}\right]\), verify that (A + 2B + 3C)^T = A^T + 2B^T + 3C^T.
Solution:
A + 2B + 3C

∴ A^T + 2B^T + 3C^T = \(\left[\begin{array}{cc}
5 & 6 \\
8 & 8 \\
2 & -2
\end{array}\right]\)
From (i) and (ii), we get
(A + 2B + 3C)^T = A^T + 2B^T + 3C^T

Question 10.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
-3 & 2 & -3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 & 1 \\
-3 & 2 \\
-1 & 3
\end{array}\right]\), prove that (A + B^T)^T = A^T + B.
prove that (A + B^T)^T = A^T + B
Solution:


From (i) and (ii), we get
(A + B^T)^T = A^T + B

Question 11.
Prove that A + A^T is a symmetric and A – A^T is a skew symmetric matrix, where
i. A = \(\left[\begin{array}{ccc}
1 & 2 & 4 \\
3 & 2 & 1 \\
-2 & -3 & 2
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
5 & 2 & -4 \\
3 & -7 & 2 \\
4 & -5 & -3
\end{array}\right]\)
Solution:

∴ (A + A^T)^T = A + A^T, i.e., A + A^T = (A + A^T)^T
∴ A + A^T is a symmetric matrix.


∴ (A – A^T)^T = – (A – A^T),
i.e., A – A^T = -(A – A^T)^T
∴ A – A^T is skew symmetric matrix.

∴ (A + A^T)^T = A + A^T, i.e., A + A^T = (A + A^T)^T
∴ A + A^T is a symmetric matrix.

∴ (A – A^T)^T = – (A – A^T),
i.e., A – A^T = -(A – A^T)^T
∴ A – A^T is skew symmetric matrix.

Question 12.
Express the following matrices as the sum of a symmetric and a skew symmetric matrix.
i. \(\left[\begin{array}{cc}
4 & -2 \\
3 & -5
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]\)
Solution:
A square matrix A can be expressed as the sum of a symmetric and a skew symmetric matrix as
A = \(\frac{1}{2}\) (A + A^T) + \(\frac{1}{2}\) (A – A^T)

P is symmetric matrix …[∵ a_ij = a_ji]
and Q is a skew symmetric matrix [∵ -a_ij = -a_ji]
A = P + Q
A = \(\left[\begin{array}{cc}
4 & \frac{1}{2} \\
\frac{1}{2} & -5
\end{array}\right]+\left[\begin{array}{ll}
0 & \frac{-5}{2} \\
\frac{5}{2} & 0
\end{array}\right]\)

∴ P is symmetric matrix …[∵ a_ij = a_ji]
and Q is a skew symmetric matrix [∵ -a_ij = -a_ji]
∴ A = P + Q
∴ A = \(\left[\begin{array}{cc}
4 & \frac{1}{2} \\
\frac{1}{2} & -5
\end{array}\right]+\left[\begin{array}{ll}
0 & \frac{-5}{2} \\
\frac{5}{2} & 0
\end{array}\right]\)

Question 13.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
3 & -2 \\
4 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
0 & 3 & -4 \\
2 & -1 & 1
\end{array}\right]\), verify that
i. (AB)^T = B^TA^T
ii. (BA)^T = A^TB^T
Solution:


From (i) and (ii), we get
(AB)^T = B^TA^T

From (i) and (ii) we get
(BA)^T = A^TB^T

Question 14.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that A^TA = I, where I is the unit matrix of order 2.
Solution:


∴ A^TA = I, where I is the unit matrix of order 2.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.5

Question 1.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 2 \\
2 & 2 \\
0 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 3 \\
-1 & 4 \\
-2 & 1
\end{array}\right]\)
Show that
i. A+B=B+A
ii. (A + B) + C = A + (B + C)
Solution:

From (i) and (ii), we get
A + B = B + A

Question 2.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & -3 \\
4 & -7
\end{array}\right]\) then find the matrix A – 2B + 6I, where I is the unit matrix of order 2.
Solution:

Question 3.
If A = \(\), B = \(\) then find the matrix C such that A + B + C is a zero matrix.
Solution:
A+ B + C is a zero matrix.
∴ A + B + C = O
C = -(A + B)

Question 4.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
3 & -5 \\
-6 & 0
\end{array}\right]\) B = \(\left[\begin{array}{cc}
-1 & -2 \\
4 & 2 \\
1 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
2 & 4 \\
-1 & -4 \\
-3 & 6
\end{array}\right]\) , find the matrix X such that 3A – 4B + 5X = C.
Solution:
3A-4B + 5X = C
∴ 5X = C + 4B – 3A

Question 5.
Solve the following equations for X and Y, if 3X – Y = \(=\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\) and X – 3Y = \(\left[\begin{array}{ll}
0 & -1 \\
0 & -1
\end{array}\right]\)
Solution:
Given equations are
\(=\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\)……………….. (i)
and X – 3Y = \(\left[\begin{array}{ll}
0 & -1 \\
0 & -1
\end{array}\right]\) ………………(ii)
By (i) x 3 – (ii) we get

Question 6.
Find the matrices A and B, if 2A – B = \(=\left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) and A – 2B = \(\left[\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right]\)
Solution:
Given equations are
2A – B = \(=\left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) ……………….. (i)
and A – 2B = \(\left[\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right]\) ……………….(ii)
By (i) – (ii) x 2, we get

Question 7.
Simplify \(\cos \theta\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\)
Solution:

Quesiton 8.
If A = \(\left[\begin{array}{cc}
1 & 2 i \\
-3 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 i & 1 \\
2 & -3
\end{array}\right]\) where i =\(\sqrt{-1}\), find A + B and A – B. Show that A + B is singular. Is A – B singular? Justify your answer.
Solution:

Question 9.
Find x and y, if \(\left[\begin{array}{ccc}
2 x+y & -1 & 1 \\
3 & 4 y & 4
\end{array}\right]+\left[\begin{array}{ccc}
-1 & 6 & 4 \\
3 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]\)
Solution:

∴ By equality of matrices, we get
2x + y – 1 = 3 and 4y = 18
∴ 2x + y = 4 and y = \(\frac{18}{4}=\frac{9}{2}\)
∴ 2x + \(\frac{9}{2}\) = 4
∴ 2x = 4 – \(\frac{9}{2}\)
∴ 2x = \(\frac{1}{2}=\)
∴ x = –\(\frac{1}{4}=\) and y = \(\frac{9}{2}=\)

Question 10.
If \(\left[\begin{array}{ll}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\), find a, b, c and d.
Solution:
\(\left[\begin{array}{ll}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\)

∴ By equality of matrices, we get
2a + b = 2 ….(i)
3a – b = 3 ….(ii)
c + 2d = 4 ….(iii)
2c – d = -1 ….(iv)
Adding (i) and (ii), we get
5a = 5
∴ a = 1
Substituting a = 1 in (i), we get
2(1) + b = 2
∴ b = 0
By (iii) + (iv) x 2, we get
5c = 2
∴ c = \(\frac{2}{5}\)
Substituting c = \(\frac{2}{5}\) in (iii), we get
\(\frac{2}{5}\) + 2d = 4
∴ 2d = 4 – \(\frac{2}{5}\)
∴ 2d = \(\frac{18}{5}\)
∴ d = \(\frac{9}{5}\)
[Note: Answer given in the textbook is d = \(\frac{3}{5}\).
However, as per our calculation it is d = \(\frac{9}{5}\).]

Question 11.
There are two book shops owned by Suresh and Ganesh. Their sales (in Rupees) for books in three subjects – Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B.

i. Find the increase in sales in Rupees from July to August 2017.
ii. If both book shops got 10% profit in the month of August 2017, find the profit for each bookseller in each subject in that month.
Solution:
i. Increase in sales in rupees from July to August 2017
For Suresh:
Increase in sales for Physics books
= 6650 – 5600= ₹ 1050
Increase in sales for Chemistry books
= 7055 – 6750 = ₹ 305
Increase in sales for Mathematics books
= 8905 – 8500 = ₹ 405
For Ganesh:
Increase in sales for Physics books
= 7000 – 6650 = ₹ 350
Increase in sales for Chemistry books
= 7500 – 7055 = ₹ 445
Increase in sales for Mathematics books
= 10200 – 8905 = ₹ 1295
[Note: Answers given in the textbook are 1760, 2090. However, as per our calculation they are 1050, 305, 405, 350, 445, 1295.]

ii. Both book shops got 10% profit in the month of August 2017.
For Suresh:
Profit for Physics books = \(\frac{6650 \times 10}{100}\) = ₹ 665
Profit for Chemistry books = \(\frac{7055 \times 10}{100}\) = ₹ 705.50
Profit for Mathematics books = \(\frac{8905 \times 10}{100}\) = ₹ 890.50

For Ganesh:
Profit for Physics books = \(\frac{7000 \times 10}{100}\) = ₹ 700
Profit for Chemistry books = \(\frac{7500 \times 10}{100}\) = ₹ 750
Profit for Mathematics books = \(\frac{10200 \times 10}{100}\) = ₹ 1020
[Note: Answers given in the textbook for Suresh’s profit in Chemistry and Mathematics books are ? 675 and ?850 respectively. However, as per our calculation profit amounts are ₹ 705.50 and ₹ 890.50 respectively.]

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.4

Question 1.
Construct a matrix A = [a_ij]_{3 x 2} whose elements ay are given by
i. a_ij = \(\frac{(\mathbf{i}-\mathbf{j})^{2}}{5-\mathbf{i}}\)
ii. a_ij = i – 3j
iii. a_ij \(\frac{(i+j)^{3}}{5}\)
Solution:

[Note: Answer given in the textbook is A = \(\left[\begin{array}{ll}
0 & \frac{1}{4} \\
\frac{1}{2} & 0 \\
2 & \frac{1}{2}
\end{array}\right]\)
However, as per our calculation it is \(\left[\begin{array}{ll}
0 & \frac{1}{4} \\
\frac{1}{3} & 0 \\
2 & \frac{1}{2}
\end{array}\right]\) ].

ii. a_ij = i – 3j
∴ a₁₁ = 1 – 3(1) = 1 – 3 = -2,
a₁₂= 1 – 3(2) = 1 – 6 = -5,
a₂₁ = 2 – 3(1) = 2 – 3 =-1,
a₂₂ = 2 – 3(2) = 2 – 6 = – 4
a₃₁ = 3 – 3(1) = 3-3 = 0,
a₃₂ = 3 – 3(2) = 3 – 6 = -3
∴ A = \(\left[\begin{array}{cc}
-2 & -5 \\
-1 & -4 \\
0 & -3
\end{array}\right]\)

iii. a_ij = \(\frac{(i+j)^{3}}{5}\)

Question 2.
Classify the following matrices as a row, a column, a square, a diagonal, a scalar, a unit, an upper triangular, a lower triangular, a symmetric or a skew- symmetric matrix.
i. \(\left[\begin{array}{ccc}
3 & -2 & 4 \\
0 & 0 & -5 \\
0 & 0 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
3 & -2 & 4 \\
0 & 0 & -5 \\
0 & 0 & 0
\end{array}\right]\)
As every element below the diagonal is zero in matrix A.
∴ A is an upper triangular matrix.

ii. \(\left[\begin{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{ccc}
0 & -4 & -7 \\
4 & 0 & 3 \\
7 & -3 & 0
\end{array}\right]\)
∴ A^T = \(-\left[\begin{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
\end{array}\right]\)
∴ A^T = -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.

iii. \(\left[\begin{array}{c}
5 \\
4 \\
-3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{c}
5 \\
4 \\
-3
\end{array}\right]\)
∴ As matrix A has only one column.
∴ A is a column matrix.

iv. \(\left[\begin{array}{lll}
9 & \sqrt{2} & -3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
9 & \sqrt{2} & -3
\end{array}\right]\)
As matrix A has only one row.
∴ A is a row matrix.

v. \(\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\)
As matrix A has all its non-diagonal elements zero and diagonal elements same.
∴ A is a scalar matrix.

vi. \(\left[\begin{array}{ccc}
2 & 0 & 0 \\
3 & -1 & 0 \\
-7 & 3 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
2 & 0 & 0 \\
3 & -1 & 0 \\
-7 & 3 & 1
\end{array}\right]\)
As every element above the diagonal is zero in matrix A.
∴ A is a lower triangular matrix.

vii. \(\left[\begin{array}{ccc}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & \frac{1}{3}
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & \frac{1}{3}
\end{array}\right]\)
As matrix A has all its non-diagonal elements zero.
∴ A is a diagonal matrix.

viii. \(\left[\begin{array}{ccc}
10 & -15 & 27 \\
-15 & 0 & \sqrt{34} \\
27 & \sqrt{34} & \frac{5}{3}
\end{array}\right]\)
Solution:

∴ A^T = A, i/e., A = A^T
∴ A is a symmetric matrix.

ix. \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
In matrix A, all the non-diagonal elements are zero and diagonal elements are one.
∴ A is a unit (identity) matrix.

x. \(\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]\)
∴ A^T = A, i/e., A = A^T
∴ A is a symmetric matrix.
∴ A is a symmetric matrix.

Question 3.
Which of the following matrices are singular or non-singular?
i. \(\left[\begin{array}{ccc}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{p} & \mathbf{q} & \mathbf{r} \\
\mathbf{2 a}-\mathbf{p} & \mathbf{2 b}-\mathbf{q} & \mathbf{2 c}-\mathbf{r}
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right]\)
iv. \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\)
Solution:

ii. Let A = \(\left[\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right|\)
Applying C₂ → C₂ + C₁
|A| = \(\left|\begin{array}{ccc}
5 & 5 & 5 \\
1 & 100 & 100 \\
6 & 105 & 105
\end{array}\right|\)
= 0 … [∵ C₂ and C₃ are identical]
∴ A is a singular matrix.

iii. Let A = \(\left[\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right| \)
= 3(5 – 8) – 5(-10 – 12) + 7(-4 – 3)
= -9 + 110 – 49 = 52 ≠ 0
∴ A is a non-singular matrix.

iv. Let A = \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\) = 49 + 20 = 69 ≠ 0

Question 4.
Find k, if the following matrices are singular.
i. \(\left[\begin{array}{cc}
7 & 3 \\
-2 & k
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
7 & 3 \\
-2 & k
\end{array}\right]\)
Since A is a singular matrix,
|A|=0
∴ \(\left|\begin{array}{cc}
7 & 3 \\
-2 & \mathrm{k}
\end{array}\right|\) = o
∴ 7k + 6 = 0
∴ 7k = -6
k = -6/7

ii. Let A = \(\left[\begin{array}{ccc}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
\end{array}\right]\)
Since A is a singular matrix,
|A|= 0
∴ \(\left|\begin{array}{ccc}
4 & 3 & 1 \\
7 & \mathrm{k} & 1 \\
10 & 9 & 1
\end{array}\right|\) = 0
∴ 4(k – 9) – 3(7 – 10) + 1(63 – 10k) = 0
∴ 4k – 36 + 9 + 63 – 10k = 0
∴ -6k + 36 = 0
∴ 6k = 36
∴ k = 6

iii. Let A = \(\left[\begin{array}{ccc}
\mathbf{k}-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]\)
Since A is a singular matrix
|A| = 0
∴ \(\left|\begin{array}{ccc}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right|\)
∴ (k – 1)(4 + 4) – 2(12 – 2) + 3 (-6 – 1) = 0
∴ 8k-8-20-21 =0
∴ 8k = 49
∴ k = 49/8

Question 5.
If A = \(\left[\begin{array}{lll}
5 & 1 & -1 \\
3 & 2 & 0
\end{array}\right]\), find (A^T)^T.
Solution:

Question 6.
If A = \(\), find (A^T)^T.
Solution:

Question 7.
Find a, b, c, if \(\left[\begin{array}{ccc}
1 & \frac{3}{5} & a \\
b & -5 & -7 \\
-4 & c & 0
\end{array}\right]\) is a symmetric matrix.
Solution:

Question 8.
Find x, y, z, if \(\) is a symmetric matrix.
Solution:

Question 9.
For each of the following matrices, using its transpose, state whether it is symmetric, skew-symmetric or neither.
i. \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
2 & 5 & 1 \\
-5 & 4 & 6 \\
-1 & -6 & 3
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
0 & 1+2 \mathbf{i} & \mathbf{i}-2 \\
-1-2 \mathbf{i} & 0 & -7 \\
2-\mathbf{i} & 7 & 0
\end{array}\right]\)
Solution:
i. Let A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
∴ A^T =\(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
∴ A^T = A, i.e., A = A^T
∴ A is a symmetric matrix.

ii.

∴ A ≠ A^T, i.e., A ≠ -A^T
∴ A is neither a symmetric nor skew-symmetric matrix.

iii.

∴ A^T = -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.

Question 10.
Construct the matrix A = [a_ij]_{3 x 3}, where a_ij = i – j. State whether A is symmetric or skew-symmetric.
Solution:
A = [a_ij]_{3 x 3}
∴ A = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
Given, a_ij = i – j
a₁₁ = 1-1 = 0, a₁₂ = 1-2 = – 1, a₁₃ = 1 – 3 = – 2,
a₂₁ – 2 – 1 = 1, a₂₂ = 2 – 2 = 0, a₂₃ =2 – 3 = – 1,
a₃₁ = 3 – 1 = 2, a₃₂ = 3 – 2 = 1, a₃₃ = 3 – 3 = 0

∴ A^T = -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.