Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.3 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3

Question 1.

Find the joint equation of the pair of lines:

(i) Through the point (2, -1) and parallel to lines represented by 2x^{2} + 3xy – 9y^{2} = 0

Solution:

The combined equation of the given lines is

2x^{2} + 3 xy – 9y^{2} = 0

i.e. 2x^{2} + 6xy – 3xy – 9y^{2} = 0

i.e. 2x(x + 3y) – 3y(x + 3y) = 0

i.e. (x + 3y)(2x – 3y) = 0

∴ their separate equations are

x + 3y = 0 and 2x – 3y = 0

∴ their slopes are m_{1} = \(\frac{-1}{3}\) and m_{2} = \(\frac{-2}{-3}=\frac{2}{3}\).

The slopes of the lines parallel to these lines are m_{1} and m_{2}, i.e. \(-\frac{1}{3}\) and \(\frac{2}{3}\).

∴ the equations of the lines with these slopes and through the point (2, -1) are

y + 1 = \(-\frac{1}{3}\) (x – 2) and y + 1 = \(\frac{2}{3}\)(x – 2)

i.e. 3y + 3= -x + 2 and 3y + 3 = 2x – 4

i.e. x + 3y + 1 = 0 and 2x – 3y – 7 = 0

∴ the joint equation of these lines is

(x + 3y + 1)(2x – 3y – 7) = 0

∴ 2x^{2} – 3xy – 7x + 6xy – 9y^{2} – 21y + 2x – 3y – 7 = 0

∴ 2x^{2} + 3xy – 9y^{2} – 5x – 24y – 7 = 0.

(ii) Through the point (2, -3) and parallel to lines represented by x^{2} + xy – y^{2} = 0

Solution:

Comparing the equation

x^{2} + xy – y^{2} = 0 … (1)

with ax^{2} + 2hxy + by^{2} = 0, we get,

a = 1, 2h = 1, b = -1

Let m_{1} and m_{2} be the slopes of the lines represented by (1).

The slopes of the lines parallel to these lines are m_{1} and m_{2}.

∴ the equations of the lines with these slopes and through the point (2, -3) are

y + 3 = m_{1}(x – 2) and y + 3 = m_{2}(x – 2)

i.e. m_{1}(x – 2) – (y + 3) = 0 and m_{2}(x – 2) – (y + 3) = 0

∴ the joint equation of these lines is

[m_{1}(x – 2) – (y + 3)][m_{2}(x – 2) – (y + 3)] = 0

∴ m_{1}m_{2}(x – 2)^{2} – m_{1}(x – 2)(y + 3) – m_{2}(x – 2)(y + 3) + (y + 3)^{2} = o

∴ m_{1}m_{2}(x – 2)^{2} – (m_{1} + m_{2})(x – 2)(y + 3) + (y + 3)^{3} = 0

∴ -(x – 2)^{2} – (x – 2)(y + 3) + (y + 3)^{2} = 0 …… [By (2)]

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

∴ (x^{2} – 4x + 4) + (xy + 3x – 2y – 6) – (y^{2} + 6y + 9) = 0

∴ x^{2} – 4x + 4 + xy + 3x – 2y – 6 – y^{2} – 6y – 9 = 0

∴ x^{2} + xy – y^{2} – x – 8y – 11 = 0.

Question 2.

Show that equation x^{2} + 2xy+ 2y^{2} + 2x + 2y + 1 = 0 does not represent a pair of lines.

Solution:

Comparing the equation

x^{2} + 2xy + 2y^{2} + 2x + 2y + 1 = 0 with

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0, we get,

a = 1, h = 1, b = 2, g = 1, f = 1, c = 1.

The given equation represents a pair of lines, if

D = \(\left|\begin{array}{lll}

a & h & g \\

h & b & f \\

g & f & c

\end{array}\right|\) = 0 and h^{2} – ab ≥ 0

Now, D = \(\left|\begin{array}{lll}

a & h & g \\

h & b & f \\

g & f & c

\end{array}\right|=\left|\begin{array}{lll}

1 & 1 & 1 \\

1 & 2 & 1 \\

1 & 1 & 1

\end{array}\right|\)

= 1 (2 – 1) – 1(1 – 1) + 1 (1 – 2)

= 1 – 0 – 1 = 0

and h^{2} – ab = (1)^{2} – 1(2) = -1 < 0

∴ given equation does not represent a pair of lines.

Question 3.

Show that equation 2x^{2} – xy – 3y^{2} – 6x + 19y – 20 = 0 represents a pair of lines.

Solution:

Comparing the equation

2x^{2} – xy – 3y^{2} – 6x + 19y – 20 = 0

with ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0, we get,

a = 2, h = \(-\frac{1}{2}\), b = -3, g = -3, f = \(\frac{19}{2}\), c = -20.

∴ D = \(\left|\begin{array}{lll}

a & h & g \\

h & b & f \\

g & f & c

\end{array}\right|=\left|\begin{array}{rrr}

2 & -\frac{1}{2} & -3 \\

-\frac{1}{2} & -3 & \frac{19}{2} \\

-3 & \frac{19}{2} & -20

\end{array}\right|\)

Taking \(\frac{1}{2}\) common from each row, we get,

D = \(\frac{1}{8}\left|\begin{array}{rrr}

4 & -1 & -6 \\

-1 & -6 & 19 \\

-6 & 19 & -40

\end{array}\right|\)

= \(\frac{1}{8}\)[4(240 – 361) + 1(40 + 114) – 6(-19 – 36)]

= \(\frac{1}{8}\)[4(-121) + 154 – 6(-55)]

= \(\frac{11}{8}\)[4(-11) + 14 – 6(-5)]

= \(\frac{1}{8}\)(-44 + 14 + 30) = 0

Also h^{2} – ab = \(\left(-\frac{1}{2}\right)^{2}\) – 2(-3) = \(\frac{1}{4}\) + 6 = \(\frac{25}{4}\) > 0

∴ the given equation represents a pair of lines.

Question 4.

Show the equation 2x^{2} + xy – y^{2} + x + 4y – 3 = 0 represents a pair of lines. Also find the acute angle between them.

Solution:

Comparing the equation

2x^{2} + xy — y^{2} + x + 4y — 3 = 0 with

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c — 0, we get,

a = 2, h = \(\frac{1}{2}\), b = -1, g = \(\frac{1}{2}\), f = 2, c = – 3.

∴ D = \(\left|\begin{array}{lll}

a & h & g \\

h & b & f \\

g & f & c

\end{array}\right|=\left|\begin{array}{lrr}

2 & \frac{1}{2} & \frac{1}{2} \\

\frac{1}{2} & -1 & 2 \\

\frac{1}{2} & 2 & -3

\end{array}\right|\)

Taking \(\frac{1}{2}\) common from each row, we get,

D = \(\frac{1}{8}\left|\begin{array}{rrr}

4 & 1 & 1 \\

1 & -2 & 4 \\

1 & 4 & -6

\end{array}\right|\)

= \(\frac{1}{8}\)[4(12 —16) — 1( —6 — 4) + 1(4 + 2)]

= \(\frac{1}{8}\)[4( – 4) – 1(-10) + 1(6)]

= \(\frac{1}{8}\)(—16 + 10 + 6) = 0

Also, h^{2} – ab = \(\left(\frac{1}{2}\right)^{2}\) – 2(-1) = \(\frac{1}{4}\) + 2 = \(\frac{9}{4}\) > 0

∴ the given equation represents a pair of lines. Let θ be the acute angle between the lines

∴ tan θ = \(\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\)

Question 5.

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

(i) (x – 2)^{2} – 3(x – 2)(y + 1) + 2(y + 1)^{2} = 0

Solution:

(x – 2)^{2} – 3(x – 2)(y + 1) + 2(y + 1)2 = 0

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

∴ (x – 2) [(x – 2) – 2(y + 1)] – (y + 1)[(x – 2) – 2(y + 1)] = 0

∴ (x – 2)(x – 2 – 2y – 2) – (y + 1)(x – 2 – 2y – 2) = 0

∴ (x – 2)(x – 2y – 4) – (y + 1)(x – 2y – 4) = 0

∴ (x – 2y – 4)(x – 2 – y – 1) = 0

∴ (x – 2y – 4)(x – y – 3) = 0

∴ the separate equations of the lines are

x – 2y – 4 = 0 and x – y – 3 = 0.

Alternative Method :

(x – 2)^{2} – 3(x – 2)(y + 1) + 2(y + 1)^{2} = 0 … (1)

Put x – 2 = X and y + 1 = Y

∴ (1) becomes,

X^{2} – 3XY + 2Y^{2} = 0

∴ X^{2} – 2XY – XY + 2Y^{2} = 0

∴ X(X – 2Y) – Y(X – 2Y) = 0

∴ (X – 2Y)(X – Y) = 0

∴ the separate equations of the lines are

∴ X – 2Y = 0 and X – Y = 0

∴ (x – 2) – 2(y + 1) = 0 and (x – 2) – (y +1) = 0

∴ x – 2y – 4 = 0 and x – y – 3 = 0.

(ii) 10(x + 1)^{2} + (x + 1)( y – 2) – 3(y – 2)^{2} = 0

Solution:

10(x + 1)^{2} + (x + 1)( y – 2) – 3(y – 2)^{2} = 0 …(1)

Put x + 1 = X and y – 2 = Y

∴ (1) becomes

10x^{2} + xy – 3y^{2} = 0

10x^{2} + 6xy – 5xy – 3y^{2} = 0

2x(5x + 3y) – y(5x + 3y) = 0

(2x – y)(5x + 3y) = 0

5x + 3y = 0 and 2x – y = 0

5x + 3y = 0

5(x + 1) + 3(y – 2) = 0

5x + 5 + 3y – 6 = 0

∴ 5x + 3y – 1 = 0

2x – y = 0

2(x + 1) – (y – 2) = 0

2x + 2 – y + 2 = 0

∴ 2x – y + 4 = 0

Question 6.

Find the value of k if the following equations represent a pair of lines :

(i) 3x^{2} + 10xy + 3y^{2} + 16y + k = 0

Solution:

Comparing the given equation with

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0,

we get, a = 3, h = 5, b = 3, g = 0, f= 8, c = k.

Now, given equation represents a pair of lines.

∴ abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0

∴ (3)(3)(k) + 2(8)(0)(5) – 3(8)^{2} – 3(0)^{2} – k(5)^{2} = 0

∴ 9k + 0 – 192 – 0 – 25k = 0

∴ -16k – 192 = 0

∴ – 16k = 192

∴ k= -12.

(ii) kxy + 10x + 6y + 4 = 0

Solution:

Comparing the given equation with

ax^{2} + 2 hxy + by^{2} + 2gx + 2fy + c = 0,

we get, a = 0, h = \(\frac{k}{2}\), b = 0, g = 5, f = 3, c = 4

Now, given equation represents a pair of lines.

∴ abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0

∴ (0)(0)(4) + 2(3)(5)\(\left(\frac{k}{2}\right)\) – 0(3)^{2} – 0(5)^{2} – 4\(\left(\frac{k}{2}\right)^{2}\) = 0

∴ 0 + 15k – 0 – 0 – k^{2} = 0

∴ 15k – k^{2} = 0

∴ -k(k – 15) = 0

∴ k = 0 or k = 15.

If k = 0, then the given equation becomes

10x + 6y + 4 = 0 which does not represent a pair of lines.

∴ k ≠ o

Hence, k = 15.

(iii) x^{2} + 3xy + 2y^{2} + x – y + k = 0

Solution:

Comparing the given equation with

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0,

we get, a = 1, h = \(\frac{3}{2}\), b = 2, g = \(\frac{1}{2}\), f= \(-\frac{1}{2}\), c = k.

Now, given equation represents a pair of lines.

∴ 2(8k – 1) – 3(6k + 1) + 1(-3 – 4) = 0

∴ 16k – 2 – 18k – 3 – 7 = 0

∴ -2k – 12 = 0

∴ -2k = 12 ∴ k = -6.

Question 7.

Find p and q if the equation px^{2} – 8xy + 3y^{2} + 14x + 2y + q = 0 represents a pair of perpendicular lines.

Solution:

The given equation represents a pair of lines perpendicular to each other

∴ (coefficient of x^{2}) + (coefficient of y^{2}) = 0

∴ p + 3 = 0 p = -3

With this value of p, the given equation is

– 3x^{2} – 8xy + 3y^{2} + 14x + 2y + q = 0.

Comparing this equation with

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0, we have,

a = -3, h = -4, b = 3, g = 7, f = 1 and c = q.

D = \(\left|\begin{array}{lll}

a & h & g \\

h & b & f \\

g & f & c

\end{array}\right|=\left|\begin{array}{rrr}

-3 & -4 & 7 \\

-4 & 3 & 1 \\

7 & 1 & q

\end{array}\right|\)

= -3(3q – 1) + 4(-4q – 7) + 7(-4 – 21)

= -9q + 3 – 16q – 28 – 175

= -25q – 200= -25(q + 8)

Since the given equation represents a pair of lines, D = 0

∴ -25(q + 8) = 0 ∴ q= -8.

Hence, p = -3 and q = -8.

Question 8.

Find p and q if the equation 2x^{2} + 8xy + py^{2} + qx + 2y – 15 = 0 represents a pair of parallel lines.

Solution:

The given equation is

2x^{2} + 8xy + py^{2} + qx + 2y – 15 = 0

Comparing it with ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0, we get,

a = 2, h = 4, b = p, g = \(\frac{q}{2}\), f = 1, c = – 15

Since the lines are parallel, h^{2} = ab

∴ (4)^{2} = 2p ∴ P = 8

Since the given equation represents a pair of lines

i.e. – 242 + 240 + 2q + 2q – 2q^{2} = 0

i.e. -2q^{2} + 4q – 2 = 0

i.e. q^{2} – 2q + 1 = 0

i.e. (q – 1)^{2} = 0 ∴ q – 1 = 0 ∴ q = 1.

Hence, p = 8 and q = 1.

Question 9.

Equations of pairs of opposite sides of a parallelogram are x^{2} – 7x+ 6 = 0 and y^{2} – 14y + 40 = 0. Find the joint equation of its diagonals.

Solution:

Let ABCD be the parallelogram such that the combined equation of sides AB and CD is x^{2} – 7x + 6 = 0 and the combined equation of sides BC and AD is y^{2} – 14y + 40 = 0.

The separate equations of the lines represented by x^{2} – 7x + 6 = 0, i.e. (x – 1)(x – 6) = 0 are x – 1 = 0 and x – 6 = 0.

Let equation of the side AB be x – 1 = 0 and equation of side CD be x – 6 = 0.

The separate equations of the lines represented by y^{2} – 14y + 40 = 0, i.e. (y – 4)(y – 10) = 0 are y – 4 = 0 and y – 10 = 0.

Let equation of the side BC be y – 4 = 0 and equation of side AD be y – 10 = 0.

Coordinates of the vertices of the parallelogram are A(1, 10), B(1, 4), C(6, 4) and D(6, 10).

∴ equation of the diagonal AC is

\(\frac{y-10}{x-1}\) = \(\frac{10-4}{1-6}\) = \(\frac{6}{-5}\)

∴ -5y + 50 = 6x – 6

∴ 6x + 5y – 56 = 0

and equation of the diagonal BD is

\(\frac{y-4}{x-1}\) = \(\frac{4-10}{1-6}\) = \(\frac{-6}{-5}\) = \(\frac{6}{5}\)

∴ 5y – 20 = 6x – 6

∴ 6x – 5y + 14 = 0

Hence, the equations of the diagonals are 6x + 5y – 56 = 0 and 6x – 5y + 14 = 0.

∴ the joint equation of the diagonals is (6x + 5y – 56)(6x – 5y + 14) = 0

∴ 36x^{2} – 30xy + 84x + 30xy – 25y^{2} + 70y – 336x + 280y – 784 = 0

∴ 36x^{2} – 25y^{2} – 252x + 350y – 784 = 0.

Question 10.

∆OAB is formed by lines x^{2} – 4xy + y^{2} = 0 and the line 2x + 3y – 1 = 0. Find the equation of the median of the triangle drawn from O.

Solution:

Let D be the midpoint of seg AB where A is (x_{1}, y_{1}) and B is (x_{2}, y_{2}).

Then D has coordinates \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\).

The joint (combined) equation of the lines OA and OB is x^{2} – 4xy + y^{2} = 0 and the equation of the line AB is 2x + 3y – 1 = 0.

∴ points A and B satisfy the equations 2x + 3y – 1 = 0

and x^{2} – 4xy + y^{2} = 0 simultaneously.

We eliminate x from the above equations, i.e.,

put x = \(\frac{1-3 y}{2}\) in the equation x^{2} – 4xy + y^{2} = 0, we get,

∴ \(\left(\frac{1-3 y}{2}\right)^{2}\) – 4\(\left(\frac{1-3 y}{2}\right)\)y + y^{2} = 0

∴ (1 – 3y)^{2} – 8(1 – 3y)y + 4y^{2} = 0

∴1 – 6y + 9y^{2} – 8y + 24y^{2} + 4y^{2} = 0

∴ 37y^{2} – 14y + 1 = 0

The roots y_{1} and y_{2} of the above quadratic equation are the y-coordinates of the points A and B.

∴ y_{1} + y_{2} = \(\frac{-b}{a}=\frac{14}{37}\)

∴ y-coordinate of D = \(\frac{y_{1}+y_{2}}{2}=\frac{7}{37}\).

Since D lies on the line AB, we can find the x-coordinate of D as

2x + 3\(\left(\frac{7}{37}\right)\) – 1 = 0

∴ 2x = 1 – \(\frac{21}{37}=\frac{16}{37}\)

∴ x = \(\frac{8}{37}\)

∴ D is (8/37, 7/37)

∴ equation of the median OD is \(\frac{x}{8 / 37}=\frac{y}{7 / 37}\),

i.e., 7x – 8y = 0.

Question 11.

Find the co-ordinates of the points of intersection of the lines represented by x^{2} – y^{2} – 2x + 1 = 0.

Solution:

Consider, x^{2} – y^{2} – 2x + 1 = 0

∴ (x^{2} – 2x + 1) – y^{2} = 0

∴ (x – 1)^{2} – y^{2} = 0

∴ (x – 1 + y)(x – 1 – y) = 0

∴ (x + y – 1)(x – y – 1) = 0

∴ separate equations of the lines are

x + y – 1 = 0 and x – y +1 = 0.

To find the point of intersection of the lines, we have to solve

x + y – 1 = 0 … (1)

and x – y + 1 = 0 … (2)

Adding (1) and (2), we get,

2x = 0 ∴ x = 0

Substituting x = 0 in (1), we get,

0 + y – 1 = 0 ∴ y = 1

∴ coordinates of the point of intersection of the lines are (0, 1).