**Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4**

## Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4

**I : Choose correct alternatives.Question 1.If the equation 4x**

^{2}+ hxy + y

^{2}= 0 represents two coincident lines, then h = _________.

(A) ± 2

(B) ± 3

(C) ± 4

(D) ± 5

**Solution:**

(C) ± 4

**Question 2.If the lines represented by kx ^{2} – 3xy + 6y^{2} = 0 are perpendicular to each other then _________.**

(A) k = 6

(B) k = -6

(C) k = 3

(D) k = -3

**Solution:**

(B) k = -6

**Question 3.Auxiliary equation of 2x ^{2} + 3xy – 9y^{2} = 0 is _________.**

(A) 2m

^{2}+ 3m – 9 = 0

(B) 9m

^{2}– 3m – 2 = 0

(C) 2m

^{2}– 3m + 9 = 0

(D) -9m

^{2}– 3m + 2 = 0

**Solution:**

(B) 9m

^{2}– 3m – 2 = 0

**Question 4.The difference between the slopes of the lines represented by 3x ^{2} – 4xy + y^{2} = 0 is _________.**

(A) 2

(B) 1

(C) 3

(D) 4

**Solution:**

(A) 2

**Question 5.If the two lines ax ^{2} +2hxy+ by^{2} = 0 make angles α and β with X-axis, then tan (α + β) = _____.**

**Question 6.**

(A) 1 : 2

(B) 2 : 1

(C) 8 : 9

(D) 9 : 8

**Solution:**

(D) 9 : 8

**Question 7.The joint equation of the lines through the origin and perpendicular to the pair of lines 3x ^{2} + 4xy – 5y^{2} = 0 is _________.**

(A) 5x

^{2}+ 4xy – 3y

^{2}= 0

(B) 3x

^{2}+ 4xy – 5y

^{2}= 0

(C) 3x

^{2}– 4xy + 5y

^{2}= 0

(D) 5x

^{2}+ 4xy + 3y

^{2}= 0

**Solution:**

(A) 5x

^{2}+ 4xy – 3y

^{2}= 0

**Question 8.If acute angle between lines ax ^{2} + 2hxy + by^{2} = 0 is, π4 then 4h^{2} = _________.**

(A) a

^{2}+ 4ab + b

^{2}

(B) a

^{2}+ 6ab + b

^{2}

(C) (a + 2b)(a + 3b)

(D) (a – 2b)(2a + b)

**Solution:**

(B) a

^{2}+ 6ab + b

^{2}

**Question 9.If the equation 3x ^{2} – 8xy + qy^{2} + 2x + 14y + p = 1 represents a pair of perpendicular lines then the values of p and q are respectively _________.**

(A) -3 and -7

(B) -7 and -3

(C) 3 and 7

(D) -7 and 3

**Solution:**

(B) -7 and -3

**Question 10.The area of triangle formed by the lines x ^{2} + 4xy + y^{2} = 0 and x – y – 4 = 0 is _________.**

**Question 11.The combined equation of the co-ordinate axes is _________.**

(A) x + y = 0

(B) x y = k

(C) xy = 0

(D) x – y = k

**Solution:**

(C) xy = 0

**Question 12.If h ^{2} = ab, then slope of lines ax^{2} + 2hxy + by^{2} = 0 are in the ratio _________.**

(A) 1 : 2

(B) 2 : 1

(C) 2 : 3

(D) 1 : 1

**Solution:**

(D) 1 : 1

[Hint: If h

^{2}= ab, then lines are coincident. Therefore slopes of the lines are equal.]

**Question 13.If slope of one of the lines ax ^{2} + 2hxy + by^{2} = 0 is 5 times the slope of the other, then 5h^{2} = _________.**
(A) ab

(B) 2 ab

(C) 7 ab

(D) 9 ab

**Solution**:

(D) 9 ab

**Question 14.If distance between lines (x – 2y) ^{2} + k(x – 2y) = 0 is 3 units, then k =**

(A) ± 3

(B) ± 5√5

(C) 0

(D) ± 3√5

**Solution:**

(D) ± 3√5

[Hint: (x – 2y)

^{2}+ k(x – 2y) = 0

∴ (x – 2y)(x – 2y + k) = 0

∴ equations of the lines are x – 2y = 0 and x – 2y + k = 0 which are parallel to each other.

**II. Solve the following.Question 1.Find the joint equation of lines:(i) x – y = 0 and x + y = 0Solution:**

The joint equation of the lines x – y = 0 and

x + y = 0 is

(x – y)(x + y) = 0

∴ x

^{2}– y

^{2}= 0.

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

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

(x + y – 3)(2x + y – 1) = 0

∴ 2x

^{2}+ xy – x + 2xy + y

^{2}– y – 6x – 3y + 3 = 0

∴ 2x

^{2}+ 3xy + y

^{2}– 7x – 4y + 3 = 0.

**(iii) Passing through the origin and having slopes 2 and 3.Solution:**

We know that the equation of the line passing through the origin and having slope m is y = mx. Equations of the lines passing through the origin and having slopes 2 and 3 are y = 2x and y = 3x respectively.

i.e. their equations are

2x – y = 0 and 3x – y = 0 respectively.

∴ their joint equation is (2x – y)(3x – y) = 0

∴ 6x

^{2}– 2xy – 3xy + y

^{2}= 0

∴ 6x

^{2}– 5xy + y

^{2}= 0.

**(iv) Passing through the origin and having inclinations 60° and 120°.Solution:**

**(v) Passing through (1, 2) amd 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 (1, 2) and parallel to the coordinate axes are x = 1 and y =1

i.e. x – 1 = 0 and y – 2 0

∴ their combined equation is

(x – 1)(y – 2) = 0

∴ x(y – 2) – 1(y – 2) = 0

∴ xy – 2x – y + 2 = 0

**(vi) Passing through (3, 2) and parallel to the line x = 2 and y = 3.Solution:**

Equations of the lines passing through (3, 2) and parallel to the lines x = 2 and y = 3 are x = 3 and y = 2.

i.e. x – 3 = 0 and y – 2 = 0

∴ their joint equation is

(x – 3)(y – 2) = 0

∴ xy – 2x – 3y + 6 = 0.

**(vii) Passing through (-1, 2) and perpendicular to the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0.Solution:**

Let L

_{1}and L

_{2}be the lines passing through the origin and perpendicular to the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0 respectively.

Since the lines L

_{1}and L

_{2}pass through the point (-1, 2), their equations are

∴ (y – y

_{1}) = m(x – x

_{1})

∴ (y – 2) = 2(x + 1)

⇒ y – 1 = 2x + 2

⇒ 2x – y + 4 = 0 and

⇒ 3y – 6 = (-4)(x + 1)

⇒ 3y – 6 = -4x + 4

⇒ 4x + 3y – 6 + 4 = 0

⇒ 4x + 3y – 2 = 0

their combined equation is

∴ (2x – y + 4)(4x + 3y – 2) = 0

∴ 8x

^{2}+ 6xy – 4x – 4xy – 3y

^{2}+ 2y + 16x + 12y – 8 = 0

∴ 8x

^{2}+ 2xy + 12x – 3y

^{2}+ 14y – 8 = 0

∴ (1 – 3)x^{2} – 2xy + y^{2} = 0

∴ -2x^{2} – 2xy + y^{2} = 0

∴ 2x^{2} + 2xy – y^{2} = 0

This is the required combined equation.

**(ix) Which are at a distance of 9 units from the Y – axis.Solution:**
Equations of the lines, which are parallel to the Y-axis and at a distance of 9 units from it, are x = 9 and x = -9

i.e. x – 9 = 0 and x + 9 = 0

∴ their combined equation is

(x – 9)(x + 9) = 0

∴ x

^{2}– 81 = 0.

**(x) Passing through the point (3, 2), one of which is parallel to the line x – 2y = 2 and other is perpendicular to the line y = 3.Solution:**

Let L

_{2}be the line passes through (3, 2) and perpendicular to the line y = 3.

∴ equation of the line L

_{2}is of the form x = a.

Since L

_{2}passes through (3, 2), 3 = a

∴ equation of the line L

_{2}is x = 3, i.e. x – 3 = 0

Hence, the equations of the required lines are

x – 2y + 1 = 0 and x – 3 = 0

∴ their joint equation is

(x – 2y + 1)(x – 3) = 0

∴ x

^{2}– 2xy + x – 3x + 6y – 3 = 0

∴ x

^{2}– 2xy – 2x + 6y – 3 = 0.

(xi) Passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18.

Solution:

Let L_{1} and L_{2} be the lines passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18 respectively.

Since the lines L_{1} and L_{2} pass through the origin, their equations are

i.e. 2x – y = 0 and x – 3y = 0

∴ their combined equation is

(2x – y)(x – 3y) = 0

∴ 2x^{2} – 6xy – xy + 3y^{2} = 0

∴ 2x^{2} – 7xy + 3y^{2} = 0.

**Question 2.Show that each of the following equation represents a pair of lines.(i) x**

^{2}+ 2xy – y

^{2}= 0Solution:

Comparing the equation x

^{2}+ 2xy – y

^{2}= 0 with ax

^{2}+ 2hxy + by

^{2}= 0, we get,

a = 1, 2h = 2, i.e. h = 1 and b = -1

∴ h

^{2}– ab = (1)

^{2}– 1(-1) = 1 + 1=2 > 0

Since the equation x

^{2}+ 2xy – y

^{2}= 0 is a homogeneous equation of second degree and h

^{2}– ab > 0, the given equation represents a pair of lines which are real and distinct.

**(ii) 4x ^{2} + 4xy + y^{2} = 0**

Solution:

Comparing the equation 4x

^{2}+ 4xy + y

^{2}= 0 with ax

^{2}+ 2hxy + by

^{2}= 0, we get,

a = 4, 2h = 4, i.e. h = 2 and b = 1

∴ h

^{2}– ab = (2)

^{2}– 4(1) = 4 – 4 = 0

Since the equation 4x

^{2}+ 4xy + y

^{2}= 0 is a homogeneous equation of second degree and h

^{2}– ab = 0, the given equation represents a pair of lines which are real and coincident.

**(iii) x ^{2} – y^{2} = 0**

Solution:

Comparing the equation x

^{2}– y

^{2}= 0 with ax

^{2}+ 2hxy + by

^{2}= 0, we get,

a = 1, 2h = 0, i.e. h = 0 and b = -1

∴ h

^{2}– ab = (0)

^{2}– 1(-1) = 0 + 1 = 1 > 0

Since the equation x

^{2}– y

^{2}= 0 is a homogeneous equation of second degree and h

^{2}– ab > 0, the given equation represents a pair of lines which are real and distinct.

**(iv) x ^{2} + 7xy – 2y^{2} = 0**

Solution:

Since the equation x

^{2}+ 7xy – 2y

^{2}= 0 is a homogeneous equation of second degree and h

^{2}– ab > 0, the given equation represents a pair of lines which are real and distinct.

**Question 3.Find the separate equations of lines represented by the following equations:(i) 6x**

^{2}– 5xy – 6y

^{2}= 0Solution:

6x

^{2}– 5xy – 6y

^{2}= 0

∴ 6x

^{2}– 9xy + 4xy – 6y

^{2}= 0

∴ 3x(2x – 3y) + 2y(2x – 3y) = 0

∴ (2x – 3y)(3x + 2y) = 0

∴ the separate equations of the lines are

2x – 3y = 0 and 3x + 2y = 0.

**(ii) x ^{2} – 4y^{2} = 0**

Solution:

x

^{2}– 4y

^{2}= 0

∴ x

^{2}– (2y)

^{2}= 0

∴(x – 2y)(x + 2y) = 0

∴ the separate equations of the lines are

x – 2y = 0 and x + 2y = 0.

**(iii) 3x ^{2} – y^{2} = 0**

Solution:

**(iv) 2x ^{2} + 2xy – y^{2} = 0**

Solution:

2x

^{2}+ 2xy – y

^{2}= 0

∴ The auxiliary equation is -m

^{2}+ 2m + 2 = 0

∴ m

^{2}– 2m – 2 = 0

**Question 4.Find the joint equation of the pair of lines through the origin and perpendicular to the linesgiven by :(i) x**

^{2}+ 4xy – 5y

^{2}= 0Solution:

Comparing the equation x

^{2}+ 4xy – 5y

^{2}= 0 with ax

^{2}+ 2hxy + by

^{2}= 0, we get,

a = 1, 2h = 4, b= -5

Let m

_{1}and m

_{2}be the slopes of the lines represented by x

^{2}+ 4xy – 5y

^{2}= 0.

**(ii) 2x ^{2} – 3xy – 9y^{2} = 0**

Solution:

**(iii) x ^{2} + xy – y^{2} = 0**

Solution:

**Question 5.Find k if(i) The sum of the slopes of the lines given by 3x**

^{2}+ kxy – y

^{2}= 0 is zero.Solution:

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.

Now, m

_{1}+ m

_{2}= 0 … (Given)

∴ k = 0.

**(ii) The sum of slopes of the lines given by 2x ^{2} + kxy – 3y^{2} = 0 is equal to their product.**

Question is modified.

The sum of slopes of the lines given by x^{2} + kxy – 3y^{2} = 0 is equal to 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.

**(iii) The slope of one of the lines given by 3x ^{2} – 4xy + ky^{2} = 0 is 1.**

Solution:

The auxiliary equation of the lines given by 3x

^{2}– 4xy + ky

^{2}= 0 is km

^{2}– 4m + 3 = 0.

Given, slope of one of the lines is 1.

∴ m = 1 is the root of the auxiliary equation km

^{2}– 4m + 3 = 0.

∴ k(1)

^{2}– 4(1) + 3 = 0

∴ k – 4 + 3 = 0

∴ k = 1.

**(iv) One of the lines given by 3x ^{2} – kxy + 5y^{2} = 0 is perpendicular to the 5x + 3y = 0.**

Solution:

**(v) The slope of one of the lines given by 3x ^{2} + 4xy + ky^{2} = 0 is three times the other.**

Solution:

3x

^{2}+ 4xy + ky

^{2}= 0

∴ divide by x

^{2}

(vi) The slopes of lines given by kx^{2} + 5xy + y^{2} = 0 differ by 1.

Solution:

Comparing the equation kx^{2} + 5xy +y^{2} = 0 with ax^{2} + 2hxy + by^{2}

the slope of the line differ by (m_{1} – m_{2}) = 1 …(1)

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

(m_{1} – m_{2})^{2} = (-5)^{2} – 4(k)

(m_{1} – m_{2})^{2} = 25 – 4k

1 = 25 – 4k ..[By (1)]

4k = 24

k = 6

**(vii) One of the lines given by 6x ^{2} + kxy + y^{2} = 0 is 2x + y = 0.**

Solution:

The auxiliary equation of the lines represented by 6x

^{2}+ kxy + y

^{2}= 0 is

m

^{2}+ km + 6 = 0.

Since one of the line is 2x + y = 0 whose slope is m = -2.

∴ m = -2 is the root of the auxiliary equation m

^{2}+ km + 6 = 0.

∴ (-2)

^{2}+ k(-2) + 6 = 0

∴ 4 – 2k + 6 = 0

∴ 2k = 10 ∴ k = 5

**Question 6.Find the joint equation of the pair of lines which bisect angle between the lines given by x ^{2} + 3xy + 2y^{2} = 0Solution:**

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

∴ separate equations of the lines represented by x

^{2}+ 3xy + 2y

^{2}= 0 are x + 2y = 0 and x + y = 0.

Let P (x, y) be any point on one of the angle bisector. Since the points on the angle bisectors are equidistant from both the lines,

the distance of P (x, y) from the line x + 2y = 0

= the distance of P(x, y) from the line x + y = 0

∴ 2(x + 2y)

^{2}= 5(x + y)

^{2}

∴ 2(x

^{2}+ 4xy + 4y

^{2}) = 5(x

^{2}+ 2xy + y

^{2})

∴ 2x

^{2}+ 8xy + 8y

^{2}= 5x

^{2}+ 10xy + 5y

^{2}

∴ 3x

^{2}+ 2xy – 3y

^{2}= 0.

This is the required joint equation of the lines which bisect the angles between the lines represented by x

^{2}+ 3xy + 2y

^{2}= 0.

**Question 7.Find the joint equation of the pair of lies through the origin and making equilateral triangle with the line x = 3.Solution:**

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

∴ OA and OB make an angle of 30° and 150° with the positive direction of X-axis

**Question 8.Show that the lines x ^{2} – 4xy + y^{2} = 0 and x + y = 10 contain the sides of an equilateral triangle. Find the area of the triangle.Solution:**

We find the joint equation of the pair of lines OA and OB through origin, each making an angle of 60° with x + y = 10 whose slope is -1.

Let OA (or OB) has slope m.

∴ its equation is y = mx … (1)

∴ x

^{2}– 4xy + y\left(\frac{y}{x}\right) = 0 is the joint equation of the two lines through the origin each making an angle of 60° with x + y = 10

∴ x

^{2}– 4xy + y

^{2}= 0 and x + y = 10 form a triangle OAB which is equilateral.

Let seg OM ⊥

^{r}line AB whose question is x + y = 10

Question 9.

If the slope of one of the lines represented by ax^{2} + 2hxy + by^{2} = 0 is three times the other then prove that 3h^{2} = 4ab.

Solution:

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

**Question 10.Find the combined equation of the bisectors of the angles between the lines represented by 5x ^{2} + 6xy – y^{2} = 0.Solution:**

Comparing the equation 5x

^{2}+ 6xy – y

^{2}= 0 with ax

^{2}+ 2hxy + by

^{2}= 0, we get,

a = 5, 2h = 6, b = -1

Let m

_{1}and m

_{2}be the slopes of the lines represented by 5x

^{2}+ 6xy – y

^{2}= 0.

The separate equations of the lines are

y = m

_{1}x and y = m

_{2}x, where m1 ≠ m2

i.e. m

_{1}x – y = 0 and m

_{1}x – y = 0.

Let P (x, y) be any point on one of the bisector of the angles between the lines.

∴ the distance of P from the line m

_{1}x – y = 0 is equal to the distance of P from the line m

_{2}x – y = 0.

∴ (m

_{2}

^{2}+ 1)(m

_{1}x – y)

^{2}= (m

_{1}

^{2}+ 1)(m

_{2}x – y)

^{2}

∴ (m

_{2}

^{2}+ 1)(m

_{1}

^{2}x

^{2}– 2m

_{1}xy + y

^{2}) = (m

_{1}

^{2}+ 1)(m

_{2}

^{2}x

^{2}– 2m

_{2}xy + y

^{2})

∴ m

_{1}

^{2}m

_{2}

^{2}x

^{2}– 2m

_{1}m

_{1}

^{2}y

^{2}xy + m

_{2}

^{2}y

^{2}+ m

_{1}

^{2}x

^{2}– 2m

_{1}

^{2}xy + y

^{2}

= m

_{1}

^{2}m

_{2}

^{2}x

^{2}– 2m

_{1}

^{2}m

_{2}xy + m

_{1}

^{2}y

^{2}+ m

_{2}

^{2}x

^{2}– 2m

_{2}xy + y

^{2}

∴ (m

_{1}

^{2}– m

_{2}

^{2})x

^{2}+ 2m

_{1}m

_{2}(m

_{1}– m

_{2})xy – 2(m

_{1}– m

_{2})xy – (m

_{1}

^{2}– m

_{2}

^{2})y

^{2}= 0

Dividing throughout by m

_{1}– m

_{2}(≠0), we get,

(m

_{1}+ m

_{2})x

^{2}+ 2m

_{1}m

_{2}xy – 2xy – (m

_{1}+ m

_{2})y

^{2}= 0

∴ 6x

^{2}– 10xy – 2xy – 6y

^{2}= 0 …[By (1)]

∴ 6x

^{2}– 12xy – 6y

^{2}= 0

∴ x

^{2}– 2xy – y

^{2}= 0

This is the joint equation of the bisectors of the angles between the lines represented by 5x

^{2}+ 6xy – y

^{2}= 0.

**Question 11.Find a, if the sum of the slopes of the lines represented by ax ^{2} + 8xy + 5y^{2} = 0 is twice their product.Solution :**

Comparing the equation ax

^{2}+ 8xy + 5y

^{2}= 0 with ax

^{2}+ 2hxy + by

^{2}= 0,

we get, a = a, 2h = 8, b = 5

Let m

_{1}and m

_{2}be the slopes of the lines represented by ax

^{2}+ 8xy + 5y

^{2}= 0.

**Question 12.If the line 4x – 5y = 0 coincides with one of the lines given by ax ^{2} + 2hxy + by^{2} = 0, then show that 25a + 40h +16b = 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.

**Question 13.Show that the following equations represent a pair of lines. Find the acute angle between them :(i) 9x**

^{2}– 6xy + y

^{2}+ 18x – 6y + 8 = 0Solution:

Comparing this equation with

ax

^{2}+ 2hxy + by

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

a = 9, h = -3, b = 1, g = 9, f = -3 and c = 8.

= 9(8 – 9) + 3(-24 + 27) + 9(9 – 9)

= 9(-1) + 3(3) + 9(0)

= -9 + 9 + 0 = 0

and h

^{2}– ab = (-3)

^{2}– 9(1) = 9 – 9 = 0

∴ the given equation represents a pair of lines.

Let θ be the acute angle between the lines.

∴ tan θ = tan0°

∴ θ = 0°.

**(ii) 2x ^{2} + xy – y^{2} + x + 4y – 3 = 0**
Comparing this equation with

Solution:

ax

^{2}+ 2hxy + by

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

a = 2, h = 1/2, b = -1, g = 1/2, f = 2 and c = -3

= -2 + 1 + 1

= -2 + 2= 0

∴ the given equation represents a pair of lines.

Let θ be the acute angle between the lines.

∴ tan θ = tan 3

∴ θ = tan

^{-1}(3)

**(iii) (x – 3) ^{2} + (x – 3)(y – 4) – 2(y – 4)^{2} = 0.**

Solution :

Hence, it represents a pair of lines passing through the new origin (3, 4).

Let θ be the acute angle between the lines.

∴ tanθ = 3 ∴ θ = tan

^{-1}(3)

**Question 14.Find the combined equation of pair of lines through the origin each of which makes angle of 60° with the Y-axis.Solution:**

Let OA and OB be the lines through the origin making an angle of 60° with the Y-axis.

Then OA and OB make an angle of 30° and 150° with the positive direction of X-axis.

**Question 15.If lines representedby ax ^{2} + 2hxy + by^{2} = 0 make angles of equal measures with the co-ordinateaxes then show that a = ± b.ORShow that, one of the lines represented by ax^{2} + 2hxy + by^{2} = 0 will make an angle of the same measure with the X-axis as the other makes with the Y-axis, if a = ± b.Solution:**

Let OA and OB be the two lines through the origin represented by ax

^{2}+ 2hxy + by

^{2}= 0.

Since these lines make angles of equal measure with the coordinate axes, they make angles ∝ and π/2 – ∝ with the positive direction of X-axis or ∝ and π/2 + ∝ with thepositive direction of X-axis.

∴ slope of the line OA = m

_{1}= tan ∝

and slope of the line OB = m

_{2}

= tan(π/2 – ∝) or tan(π/2 + ∝)

i.e. m

_{2}= cot ∝ or m

_{2}= -cot ∝

∴ m

_{1}m

_{2}– tan ∝ x cot ∝ = 1

OR m

_{1}m

_{2}= tan ∝ (-cot ∝) = -1

i.e. m

_{1}m

_{2}= ± 1

But m

_{1}m

_{2}= a/b

∴ a/b= ±1 ∴ a = ±b

This is the required condition.

**Question 16.Show that the combined equation of a pair of lines through the origin and each making an angle of ∝ with the line x + y = 0 is x ^{2} + 2(sec 2∝) xy + y^{2} = 0.Solution:**

Let OA and OB be the required lines.

Let OA (or OB) has slope m.

∴ its equation is y = mx … (1)

It makes an angle ∝ with x + y = 0 whose slope is -1. m +1

∴ tan

^{2}∝(1 – 2m + m

^{2}) = m

^{2}+ 2m + 1

∴ tan

^{2}∝ – 2m tan

^{2}∝ + m

^{2}tan

^{2}∝ = m

^{2}+ 2m + 1

∴ (tan

^{2}∝ – 1)m

^{2}– 2(1 + tan

^{2}∝)m + (tan

^{2}∝ – 1) = 0

∴ y

^{2}+ 2xysec2∝ + x

^{2}= 0

∴ x

^{2}+ 2(sec2∝)xy + y

^{2}= 0 is the required equation.

**Question 17.Show that the line 3x + 4y+ 5 = 0 and the lines (3x + 4y) ^{2} – 3(4x – 3y)^{2} =0 form an equilateral triangle.Solution:**

Let m be the slope of one of the line making an angle of 60° with the line 3x + 4y + 5 = 0. The angle between the lines having slope m and m

_{1}is 60°.

On squaring both sides, we get,

∴ 3 (4 – 3m)

^{2}= (4m + 3)

^{2}

∴ 3(16 – 24m + 9m

^{2}) = 16m

^{2}+ 24m + 9

∴ 48 – 72m + 27m

^{2}= 16m

^{2}+ 24m + 9

∴ 11m

^{2}– 96m + 39 = 0

This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = y/x.

∴ the combined equation of the two lines is

∴ 11y

^{2}– 96xy + 39x

^{2}= 0

∴ 39x

^{2}– 96xy + 11y

^{2}= 0.

∴ 39x

^{2}– 96xy + 11y

^{2}= 0 is the joint equation of the two lines through the origin each making an angle of 60° with the line 3x + 4y + 5 = 0.

The equation 39x

^{2}– 96xy + 11y

^{2}= 0 can be written as :

-39x

^{2}+ 96xy – 11y

^{2}= 0

i.e., (9x

^{2}– 48x

^{2}) + (24xy + 72xy) + (16y

^{2}– 27y

^{2}) = 0

i.e. (9x

^{2}+ 24xy + 16y

^{2}) – (48x

^{2}– 72xy + 27y

^{2}) = 0

i.e. (9x

^{2}+ 24xy + 16y

^{2}) – 3(16x

^{2}– 24xy + 9y

^{2}) = 0

i.e. (3x + 4y)

^{2}– 3(4x – 3y)

^{2}= 0

Hence, the line 3x + 4y + 5 = 0 and the lines

(3x + 4y)

^{2}– 3(4x – 3y)

^{2}form the sides of an equilateral triangle.

**Question 18.Show that lines x ^{2} – 4xy + y^{2} = 0 and x + y = √6 form an equilateral triangle. Find its area and perimeter.Solution:**

x

^{2}– 4xy + y

^{2}= 0 and x + y = √6 form a triangle OAB which is equilateral.

Let OM be the perpendicular from the origin O to AB whose equation is x + y = √6

In right angled triangle OAM,

∴ OA = 2

∴ length of the each side of the equilateral triangle OAB = 2 units.

∴ perimeter of ∆ OAB = 3 × length of each side

= 3 × 2 = 6 units.

**Question 19.If the slope of one of the lines given by ax ^{2} + 2hxy + by^{2} = 0 is square of the other then show that a^{2}b + ab^{2} + 8h^{3} = 6abh.Solution:**

Let m be the slope of one of the lines given by ax

^{2}+ 2hxy + by

^{2}= 0.

Then the other line has slope m

^{2}

Multiplying by b

^{3}, we get,

-8h

^{3}= ab

^{2}+ a

^{2}b – 6abh

∴ a

^{2}b + ab

^{2}+ 8h

^{3}= 6abh

This is the required condition.

**Question 20.**

**Question 21.Show that the difference between the slopes of lines given by (tan ^{2}θ + cos^{2}θ )x^{2} – 2xytanθ + (sin^{2}θ )y^{2} = 0 is two.Solution:**

Comparing the equation (tan

^{2}θ + cos

^{2}θ)x

^{2}– 2xy tan θ + (sin

^{2}θ) y

^{2}= 0 with ax

^{2}+ 2hxy + by

^{2}= 0, we get,

a = tan

^{2}θ + cos

^{2}θ, 2h = -2 tan θ and b = sin2θ

Let m

_{1}and m

_{2}be the slopes of the lines represented by the given equation.

**Question 22.Find the condition that the equation ay ^{2} + bxy + ex + dy = 0 may represent a pair of lines.Solution:**

Comparing the equation

ay

^{2}+ bxy + ex + dy = 0 with

Ax

^{2}+ 2Hxy + By

^{2}+ 2Gx + 2Fy + C = 0, we get,

A = 0, H = b2, B = a,G = e2, F = d2, C = 0

The given equation represents a pair of lines,

i.e. if bed – ae

^{2}= 0

i.e. if e(bd – ae) = 0

i.e. e = 0 or bd – ae = 0

i.e. e = 0 or bd = ae

This is the required condition.

**Question 23.If the lines given by ax ^{2} + 2hxy + by^{2} = 0 form an equilateral triangle with the line lx + my = 1 then show that (3a + b)(a + 3b) = 4h^{2}.Solution:**
Since the lines ax

^{2}+ 2hxy + by

^{2}= 0 form an equilateral triangle with the line lx + my = 1, the angle between the lines ax

^{2}+ 2hxy + by

^{2}= 0 is 60°.

∴ 3(a + b)

^{2}= 4(h

^{2}– ab)

∴ 3(a

^{2}+ 2ab + b

^{2}) = 4h

^{2}– 4ab

∴ 3a

^{2}+ 6ab + 3b

^{2}+ 4ab = 4h

^{2}

∴ 3a

^{2}+ 10ab + 3b

^{2}= 4h

^{2}

∴ 3a

^{2}+ 9ab + ab + 3b

^{2}= 4h

^{2}

∴ 3a(a + 3b) + b(a + 3b) = 4h

^{2}

∴ (3a + b)(a + 3b) = 4h

^{2}

This is the required condition.

**Question 24.If line x + 2 = 0 coincides with one of the lines represented by the equation x ^{2} + 2xy + 4y + k = 0 then show that k = -4.Solution:**

One of the lines represented by

x

^{2}+ 2xy + 4y + k = 0 … (1)

is x + 2 = 0.

Let the other line represented by (1) be ax + by + c = 0.

∴ their combined equation is (x + 2)(ax + by + c) = 0

∴ ax

^{2}+ bxy + cx + 2ax + 2by + 2c = 0

∴ ax

^{2}+ bxy + (2a + c)x + 2by + 2c — 0 … (2)

As the equations (1) and (2) are the combined equations of the same two lines, they are identical.

∴ by comparing their corresponding coefficients, we get,

∴ 1 = −4k

∴ k = -4.

**Question 25.Prove that the combined equation of the pair of lines passing through the origin and perpendicular to the lines represented by ax ^{2} + 2hxy + by^{2} = 0 is bx^{2} – 2hxy + ay^{2} = 0Solution:**
Let m

_{1}and m

_{2}be the slopes of the lines represented by ax

^{2}+ 2hxy + by

^{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

∴ bx

^{2}– 2hxy + ay

^{2}= 0.

**Question 26.If equation ax ^{2} – y^{2} + 2y + c = 1 represents a pair of perpendicular lines then find a and c.Solution:**

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

∴ coefficient of x

^{2}+ coefficient of y

^{2}= 0

∴ a – 1 = 0 ∴ a = 1

With this value of a, the given equation is

x

^{2}– y

^{2}+ 2y + c – 1 = 0

Comparing this equation with

Ax

^{2}+ 2Hxy + By

^{2}+ 2Gx + 2Fy + C = 0, we get,

A = 1, H = 0, B = -1, G = 0, F = 1, C = c – 1

Since the given equation represents a pair of lines,

∴ 1(-c + 1 – 1) – 0 + 0 = 0

∴ -c = 0

∴ c = 0.

Hence, a = 1, c = 0.