Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.1 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.1

Question 1.

Find the combined equation of the following pairs of lines:

(i) 2x + y = 0 and 3x – y = 0

Solution:

The combined equation of the lines 2x + y = 0 and 3x – y = 0 is

(2x + y)( 3x – y) = 0

∴ 6x^{2} – 2xy + 3xy – y^{2} = 0

∴ 6x^{2} – xy – y^{2} = 0.

(ii) x + 2y – 1 = 0 and x – 3y + 2 = 0

Solution:

The combined equation of the lines x + 2y – 1 = 0 and x – 3y + 2 = 0 is

(x + 2y – 1)(x – 3y + 2) = 0

∴ x^{2} – 3xy + 2x + 2xy – 6y^{2} + 4y – x + 3y – 2 = 0

∴ x^{2} – xy – 6y^{2} + x + 7y – 2 = 0.

(iii) Passing through (2, 3) and parallel to the co-ordinate axes.

Solution:

Equations of the coordinate axes are x = 0 and y = 0.

∴ the equations of the lines passing through (2, 3) and parallel to the coordinate axes are x = 2 and

i.e. x – 2 = 0 and y – 3 = 0.

∴ their combined equation is

(x – 2)(y – 3) = 0.

∴ xy – 3x – 2y + 6 = 0.

(iv) Passing through (2, 3) and perpendicular to lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0

Solution:

Let L_{1} and L_{2} be the lines passing through the point (2, 3) and perpendicular to the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 respectively.

Slopes of the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 are \(\frac{-3}{2}\) and \(\frac{-1}{-3}=\frac{1}{3}\) respectively.

∴ slopes of the lines L_{1} and L_{2} are \(\frac{2}{3}\) and -3 respectively.

Since the lines L_{1} and L_{2} pass through the point (2, 3), their equations are

y – 3 = \(\frac{2}{3}\)(x – 2) and y – 3 = -3 (x – 2)

∴ 3y – 9 = 2x – 4 and y – 3= -3x + 6

∴ 2x – 3y + 5 = 0 and 3x – y – 9 = 0

∴ their combined equation is

(2x – 3y + 5)(3x + y – 9) = 0

∴ 6x^{2} + 2xy – 18x – 9xy – 3y^{2} + 27y + 15x + 5y – 45 = 0

∴ 6x^{2} – 7xy – 3y^{2} – 3x + 32y – 45 = 0.

(v) Passsing through (-1, 2),one is parallel to x + 3y – 1 = 0 and the other is perpendicular to 2x – 3y – 1 = 0.

Solution:

Let L_{1} be the line passing through (-1, 2) and parallel to the line x + 3y – 1 = 0 whose slope is –\(\frac{1}{3}\).

∴ slope of the line L_{1} is –\(\frac{1}{3}\)

∴ equation of the line L_{1} is

y – 2 = –\(\frac{1}{3}\)(x + 1)

∴ 3y – 6 = -x – 1

∴ x + 3y – 5 = 0

Let L_{2} be the line passing through (-1, 2) and perpendicular to the line 2x – 3y – 1 = 0

whose slope is \(\frac{-2}{-3}=\frac{2}{3}\).

∴ slope of the line L_{2} is –\(\frac{3}{2}\)

∴ equation of the line L_{2} is

y – 2= –\(\frac{3}{2}\)(x + 1)

∴ 2y – 4 = -3x – 3

∴ 3x + 2y – 1 = 0

Hence, the equations of the required lines are

x + 3y – 5 = 0 and 3x + 2y – 1 = 0

∴ their combined equation is

(x + 3y – 5)(3x + 2y – 1) = 0

∴ 3x^{2} + 2xy – x + 9xy + 6y^{2} – 3y – 15x – 10y + 5 = 0

∴ 3x^{2} + 11xy + 6y^{2} – 16x – 13y + 5 = 0

Question 2.

Find the separate equations of the lines represented by following equations:

(i) 3y^{2} + 7xy = 0

Solution:

3y^{2} + 7xy = 0

∴ y(3y + 7x) = 0

∴ the separate equations of the lines are y = 0 and 7x + 3y = 0.

(ii) 5x^{2} – 9y^{2} = 0

Solution:

5x^{2} – 9y^{2} = 0

∴ (\(\sqrt {5}\) x)^{2} – (3y)^{2} = 0

∴ (\(\sqrt {5}\)x + 3y)(\(\sqrt {5}\)x – 3y) = 0

∴ the separate equations of the lines are

\(\sqrt {5}\)x + 3y = 0 and \(\sqrt {5}\)x – 3y = 0.

(iii) x^{2} – 4xy = 0

Solution:

x^{2} – 4xy = 0

∴ x(x – 4y) = 0

∴ the separate equations of the lines are x = 0 and x – 4y = 0

(iv) 3x^{2} – 10xy – 8y^{2} = 0

Solution:

3x^{2} – 10xy – 8y^{2} = 0

∴ 3x^{2} – 12xy + 2xy – 8y^{2} = 0

∴ 3x(x – 4y) + 2y(x – 4y) = 0

∴ (x – 4y)(3x +2y) = 0

∴ the separate equations of the lines are x – 4y = 0 and 3x + 2y = 0.

(v) 3x^{2} – \(2 \sqrt{3}\) xy – 3y^{2} = 0

Solution:

3x^{2} – 2\(\sqrt {3}\)xy – 3y2 = 0

∴ 3x^{2} – 3\(\sqrt {3}\)xy + \(\sqrt {3}\)xy – 3y^{2} = 0

∴ 3x(x – \(\sqrt {3}\)y) + \(\sqrt {3}\)y(x – \(\sqrt {3}\)y) = 0

∴ (x – \(\sqrt {3}\)y)(3x + \(\sqrt {3}\)y) = 0

∴ the separate equations of the lines are

∴ x – \(\sqrt {3}\)y = 0 and 3x + \(\sqrt {3}\)y = 0.

(vi) x^{2} + 2(cosec ∝)xy + y^{2} = 0

Solution:

x^{2} + 2 (cosec ∝)xy – y^{2} = 0

i.e. y^{2} + 2(cosec∝)xy + x^{2} = 0

Dividing by x^{2}, we get,

∴ the separate equations of the lines are

(cosec ∝ – cot ∝)x + y = 0 and (cosec ∝ + cot ∝)x + y = 0.

(vii) x^{2} + 2xy tan ∝ – y^{2} = 0

Solution:

x^{2} + 2xy tan ∝ – y^{2} = 0

Dividind by y^{2}

The separate equations of the lines are

(sec∝ – tan ∝)x + y = 0 and (sec ∝ + tan ∝)x – y = 0

Question 3.

Find the combined equation of a pair of lines passing through the origin and perpendicular

to the lines represented by following equations :

(i) 5x^{2} – 8xy + 3y^{2} = 0

Solution:

Comparing the equation 5x^{2} – 8xy + 3y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 5, 2h = -8, b = 3

Let m_{1} and m_{2} be the slopes of the lines represented by 5x^{2} – 8xy + 3y^{2} = 0.

∴ m_{1} + m_{2} = \(\frac{-2 h}{b}=\frac{8}{3}\)

amd m_{1}m_{2} = \(\frac{a}{b}=\frac{5}{3}\) …(1)

Now required lines are perpendicular to these lines

∴ their slopes are -1 /m_{1} and -1/m_{2} Since these lines are passing through the origin, their separate equations are

y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x

i.e. m_{1}y = -x and m_{2}y = -x

i.e. x + m_{1}y = 0 and x + m_{2}y = 0

∴ their combined equation is

(x + m_{1}y) (x + m_{2}y) = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + \(\frac{8}{3}\)xy + \(\frac{5}{3}\)y^{2} = 0 … [By (1)]

∴ x^{2} + 8xy + 5y\(\frac{8}{3}\) = 0

(ii) 5x^{2} + 2xy – 3y^{2} = 0

Solution:

Comparing the equation 5x^{2} + 2xy – 3y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 5, 2h = 2, b = -3

Let m_{1} and m_{2} be the slopes of the lines represented by 5x^{2} + 2xy – 3y^{2} = 0

∴ m_{1} + m_{2} = \(\frac{-2 h}{b}=\frac{-2}{-3}=\frac{2}{3}\) and m_{1}m_{2} = \(\frac{a}{b}=\frac{5}{-3}\) ..(1)

Now required lines are perpendicular to these lines

∴ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\)

Since these lines are passing through the origin, their separate equations are

y = \(\frac{-1}{\mathrm{~m}_{1}}\)x and y = \(\frac{-1}{\mathrm{~m}_{2}}\)x

i.e. m_{1}y = -x amd m_{2}y = -x

i.e. x + m_{1}y = 0 and x + m_{2}y = 0

∴ their combined equation is

∴ (x + m_{1}y)(x + m_{2}y) = 0

x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} + \(\frac{2}{3}\)xy – \(\frac{5}{3}\)y = 0 …[By (1)]

∴ 3x^{2} + 2xy – 5y^{2} = 0

(iii) xy + y^{2} = 0

Solution:

Comparing the equation xy + y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 0, 2h = 1, b = 1

Let m_{1} and m_{2} be the slopes of the lines represented by xy + y^{2} = 0

Now required lines are perpendicular to these lines

∴ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\).

Since these lines are passing through the origin, their separate equations are

y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x

i.e. m_{1}y = -x and m_{2}y = -x

i.e. x + m_{1}y = 0 and x + m_{2}y = 0

∴ their combined equation is

(x + m_{1}y) (x + m_{2}y) = 0

∴ x^{2} + (m_{1} + m_{2})xy + m_{1}m_{2}y^{2} = 0

∴ x^{2} – xy = 0.y^{2} = 0 … [By (1)]

∴ x^{2} – xy = 0.

Alternative Method :

Consider xy + y^{2} = 0

∴ y(x + y) = 0

∴ separate equations of the lines are y = 0 and

3x^{2} + 8xy + 5y^{2} = 0.

x + y = 0.

Let m_{1} and m_{2} be the slopes of these lines.

Then m_{1} = 0 and m_{2} = -1

Now, required lines are perpendicular to these lines.

∴ their slopes are \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)

Since, m_{1} = 0, \(-\frac{1}{m_{1}}\) does not exist.

Also, m_{2} = -1, \(-\frac{1}{m_{2}}\) = 1

Since these lines are passing through the origin, their separate equations are x = 0 and y = x,

i.e. x – y = 0

∴ their combined equation is

x(x – y) = 0

x^{2} – xy = 0.

(iv) 3x^{2} – 4xy = 0

Solution:

Consider 3x^{2} – 4xy = 0

∴ x(3x – 4y) = 0

∴ separate equations of the lines are x = 0 and 3x – 4y = 0.

Let m_{1} and m_{2} be the slopes of these lines.

Then m_{1} does not exist and and m_{1} = \(\frac{3}{4}\).

Now, required lines are perpendicular to these lines.

∴ their slopes are \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\).

Since m_{1} does not exist, \(-\frac{1}{m_{1}}\) = 0

Also m_{2} = \(\frac{3}{4^{\prime}}-\frac{1}{m_{2}}=-\frac{4}{3}\)

Since these lines are passing through the origin, their separate equations are y = 0 and y = \(-\frac{4}{3}\)x,

i.e. 4x + 3y = 0

∴ their combined equation is

y(4x + 3y) = 0

∴ 4xy + 3y^{2} = 0.

Question 4.

Find k if,

(i) the sum of the slopes of the lines represented by x^{2} + kxy – 3y^{2} = 0 is twice their product.

Solution:

Comparing the equation x^{2} + kxy – 3y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get, a = 1, 2h = k, b = -3.

Let m_{1} and m_{2} be the slopes of the lines represented by x^{2} + kxy – 3y^{2} = 0.

∴ m_{1} + m_{2} = \(\frac{-2 h}{b}=-\frac{k}{(-3)}=\frac{k}{3}\)

and m_{1}m_{2} = \(\frac{a}{b}=\frac{1}{(-3)}=-\frac{1}{3}\)

Now, m_{1} + m_{2} = 2(m_{1}m_{2}) ..(Given)

∴ \(\frac{k}{3}=2\left(-\frac{1}{3}\right)\) ∴ k = -2

(ii) slopes of lines represent by 3x^{2} + kxy – y^{2} = 0 differ by 4.

Solution:

(ii) Comparing the equation 3x^{2} + kxy – y^{2} = 0 with ax^{2} + 2hxy + by^{2} = 0, we get, a = 3, 2h = k, b = -1.

Let m_{1} and m_{2} be the slopes of the lines represented by 3x^{2} + kxy – y^{2} = 0.

∴ m_{1} + m_{2} = \(\frac{-2 h}{b}=-\frac{k}{-1}\) = k

and m_{1}_{2} = \(\frac{a}{b}=\frac{3}{-1}\) = -3

∴ (m_{1} – m_{2})^{2} = (m_{1} + m_{2})^{2} – 4m_{1}m_{2}

= k^{2} – 4 (-3)

= k^{2} + 12 … (1)

But |m_{1} – m_{2}| =4

∴ (m_{1} – m_{2})^{2} = 16 … (2)

∴ from (1) and (2), k^{2} + 12 = 16

∴ k^{2} = 4 ∴ k= ±2.

(iii) slope of one of the lines given by kx^{2} + 4xy – y^{2} = 0 exceeds the slope of the other by 8.

Solution:

Comparing the equation kx^{2} + 4xy – y^{2} = 0 with ^{2} + 2hxy + by^{2} = 0, we get, a = k, 2h = 4, b = -1. Let m_{1} and m_{2} be the slopes of the lines represented by kx^{2} + 4xy – y^{2} = 0.

∴ m_{1} + m_{2} = \(\frac{-2 h}{b}=\frac{-4}{-1}\) = 4

and m_{1}m_{2} = \(\frac{a}{b}=\frac{k}{-1}\) = -k

We are given that m_{2} = m_{1} + 8

m_{1} + m_{1} + 8 = 4

∴ 2m_{1} = -4 ∴ m_{1} = -2 … (1)

Also, m_{1}(m_{1} + 8) = -k

(-2)(-2 + 8) = -k … [By(1)]

∴ (-2)(6) = -k

∴ -12= -k ∴ k = 12.

Question 5.

Find the condition that :

(i) the line 4x + 5y = 0 coincides with one of the lines given by ax^{2} + 2hxy + by^{2} = 0.

Solution:

The auxiliary equation of the lines represented by ax^{2} + 2hxy + by^{2} = 0 is bm^{2} + 2hm + a = 0.

Given that 4x + 5y = 0 is one of the lines represented by ax^{2} + 2hxy + by^{2} = 0.

The slope of the line 4x + 5y = 0 is \(-\frac{4}{5}\).

∴ m = \(-\frac{4}{5}\) is a root of the auxiliary equation bm^{2} + 2hm + a = 0.

∴ b\(\left(-\frac{4}{5}\right)^{2}\) + 2h\(\left(-\frac{4}{5}\right)\) + a = 0

∴ \(\frac{16 b}{25}-\frac{8 h}{5}\) + a = 0

∴ 16b – 40h + 25a = 0

∴ 25a + 16b = 40k.

This is the required condition.

(ii) the line 3x + y = 0 may be perpendicular to one of the lines given by ax^{2} + 2hxy + by^{2} = 0.

Solution:

The auxiliary equation of the lines represented by ax^{2} + 2hxy + by^{2} = 0 is bm^{2} + 2hm + a = 0.

Since one line is perpendicular to the line 3x + y = 0

whose slope is \(-\frac{3}{1}\) = -3

∴ slope of that line = m = \(\frac{1}{3}\)

∴ m = \(\frac{1}{3}\)is the root of the auxiliary equation bm^{2} + 2hm + a = 0.

∴ b\(\left(\frac{1}{3}\right)^{2}\) + 2h\(\left(\frac{1}{3}\right)\) + a = 0

∴ \(\frac{b}{9}+\frac{2 h}{3}\) + a = 0

∴ b + 6h + 9a = 0

∴ 9a + b + 6h = 0

This is the required condition.

Question 6.

If one of the lines given by ax^{2} + 2hxy + by^{2} = 0 is perpendicular to px + qy = 0 then show that ap^{2} + 2hpq + bq^{2} = 0.

Solution:

To prove ap^{2} + 2hpq + bq^{2} = 0.

Let the slope of the pair of straight lines ax^{2} + 2hxy + by^{2} = 0 be m_{1} and m_{2}

Then, m_{1} + m_{2} = \(\frac{-2 h}{b}\) and m_{1}m_{2} = \(\frac{a}{b}\)

Slope of the line px + qy = 0 is \(\frac{-p}{q}\)

But one of the lines of ax^{2} + 2hxy + by^{2} = 0 is perpendicular to px + qy = 0

⇒ bq^{2} + ap^{2} = -2hpq

⇒ ap^{2} + 2hpq + bq^{2} = 0

Question 7.

Find the combined equation of the pair of lines passing through the origin and making an equilateral triangle with the line y = 3.

Solution:

Let OA and OB be the lines through the origin making.an angle of 60° with the line y = 3.

∴ OA and OB make an angle of 60° and 120° with the positive direction of X-axis.

∴ slope of OA = tan60° = \(\sqrt {3}\)

∴ equation of the line OA is

y = \(\sqrt {3}\) x, i.e. \(\sqrt {3}\) x – y = 0

Slope of OB = tan 120° = tan (180° – 60°)

= -tan 60°= –\(\sqrt {3}\)

∴ equation of the line OB is

y = –\(\sqrt {3}\) x, i.e. \(\sqrt {3}\) x + y = 0

∴ required joint equation of the lines is

(\(\sqrt {3}\) x – y)(\(\sqrt {3}\) x + y) = 0

i.e. 3x^{2} – y^{2} = 0.

Question 8.

If slope of one of the lines given by ax^{2} + 2hxy + by^{2} = 0 is four times the other then show that 16h^{2} = 25ab.

Solution:

Let m_{1} and m_{2} be the slopes of the lines given by ax^{2} + 2hxy + by^{2} = 0.

∴ m_{1} + m_{2} = \(-\frac{2 h}{b}\)

and m_{1}m_{2} = \(\frac{a}{b}\)

We are given that m_{2} = 4m_{1}

∴ 16h^{2} = 25ab

This is the required condition.

Question 9.

If one of the lines given by ax^{2} + 2hxy + by^{2} = 0 bisects an angle between co-ordinate axes then show that (a + b) ^{2} = 4h^{2}.

Solution:

The auxiliary equation of the lines given by ax^{2} + 2hxy + by^{2} = 0 is bm^{2} + 2hm + a = 0.

Since one of the line bisects an angle between the coordinate axes, that line makes an angle of 45° or 135° with the positive direction of X-axis.

∴ slope of that line = tan45° or tan 135°

∴ m = tan45° = 1

or m = tan 135° = tan (180° – 45°)

= -tan 45°= -1

∴ m = ±1 are the roots of the auxiliary equation bm^{2} + 2hm + a = 0.

∴ b(±1)^{2} + 2h(±1) + a = 0

∴ b ± 2h + a = 0

∴ a + b = ±2h

∴ (a + b)^{2} = 4h^{2}

This is the required condition.