**Chapter 4 Pair of Straight Lines Ex 4.3**

## 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}= 0Solution:

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

**(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

= 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,

**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,

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

**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}= 0Solution:

(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 = 0Solution:

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 = 0Solution:**

Comparing the given equation with

ax

^{2}+ 2 hxy + by

^{2}+ 2gx + 2fy + c = 0,

we get, a = 0, h = 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)(k/2) – 0(3)

^{2}– 0(5)

^{2}– 4(k/2)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,

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.

= -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 = 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

∴ -5y + 50 = 6x – 6

∴ 6x + 5y – 56 = 0

and equation of the diagonal BD is

∴ 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}).

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.,

∴ (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.

**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).