**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 = 0Solution:**

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 = 0Solution:**

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 = 0Solution:**

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

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 = 0Solution:

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:

**(iii) x ^{2} – 4xy = 0Solution:**

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.

**(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 perpendicularto the lines represented by following equations :(i) 5x**

^{2}– 8xy + 3y

^{2}= 0Solution:

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.

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

**(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

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

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

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

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.

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 = 0Solution:**

Consider 3x

^{2}– 4xy = 0

∴ x(3x – 4y) = 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.

**(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})

^{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.

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.

**(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

**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:**

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.

**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}= −2h/b

and m

_{1}m

_{2}= 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.