**Question 1.Show that lines represented by 3x ^{2} – 4xy – 3y^{2} = 0 are perpendicular to each other.Solution:**

Comparing the equation 3x

^{2}– 4 xy – 3y

^{2}= 0 with ax

^{2}+ 2hxy + by

^{2}= 0, we get, a = 3, 2h = -4, b = -3 Since a + b = 3 + (-3) = 0, the lines represented by 3x

^{2}– 4xy – 3y

^{2}= 0 are perpendicular to each other.

**Question 2.Show that lines represented by x ^{2} + 6xy + gy^{2}= 0 are coincident.Question is modified.Show that lines represented by x^{2} + 6xy + 9y^{2}= 0 are coincident.Solution:**

Comparing the equation x

^{2}+ 6xy + 9y

^{2}= 0 with ax

^{2}+ 2hxy + by

^{2}= 0, we get,

a = 1, 2h = 6, i.e. h = 3 and b = 9

Since h

^{2}– ab = (3)

^{2}– 1(9)

= 9 – 9 = 0, .

the lines represented by x

^{2}+ 6xy + 9y

^{2}= 0 are coincident.

**Question 3.Find the value of k if lines represented by kx ^{2} + 4xy – 4y^{2} = 0 are perpendicular to each other.Solution:**

Comparing the equation kx

^{2}+ 4xy – 4y

^{2}= 0 with ax

^{2}+ 2hxy + by

^{2}= 0, we get,

a = k, 2h = 4, b = -4

Since lines represented by kx

^{2}+ 4xy – 4y

^{2}= 0 are perpendicular to each other,

a + b = 0

∴ k – 4 = 0 ∴ k = 4.

**Question 4.Find the measure of the acute angle between the lines represented by:**

∴ θ = 30°.

**(ii) 4x ^{2} + 5xy + y^{2} = 0**

Solution:

Comparing the equation 4x

^{2}+ 5xy + y

^{2}= 0 with ax

^{2}+ 2hxy + by

^{2}= 0, we get,

a = 4, 2h = 5, i.e. h = 52 and b = 1.

Let θ be the acute angle between the lines.

**(iii) 2x ^{2} + 7xy + 3y^{2} = 0**

Solution:

Comparing the equation

2x

^{2}+ 7xy + 3y

^{2}= 0 with

ax

^{2}+ 2hxy + by

^{2}= 0, we get,

a = 2, 2h = 7 i.e. h = 7/2 and b = 3

Let θ be the acute angle between the lines.

tanθ = 1

∴ θ = tan 1 = 45°

∴ θ = 45°

**(iv) (a ^{2} – 3b^{2})x^{2} + 8abxy + (b^{2} – 3a^{2})y^{2} = 0**

Solution:

Comparing the equation

(a

^{2}– 3b

^{2})x

^{2}+ 8abxy + (b

^{2}– 3a

^{2})y

^{2}= 0, with

Ax

^{2}+ 2Hxy + By

^{2}= 0, we have,

A = a

^{2}– 3b

^{2}, H = 4ab, B = b

^{2}– 3a

^{2}.

∴ H

^{2}– AB = 16a

^{2}b

^{2}– (a

^{2}– 3b

^{2})(b

^{2}– 3a

^{2})

= 16a

^{2}b

^{2}+ (a

^{2}– 3b

^{2})(3a

^{2}– b

^{2})

= 16a

^{2}b

^{2}+ 3a

^{4}– 10a

^{2}b

^{2}+ 3b

^{4}

= 3a

^{4}+ 6a

^{2}b

^{2}+ 3b

^{4}

= 3(a

^{4}+ 2a

^{2}b

^{2}+ b

^{4})

= 3 (a

^{2}+ b

^{2})

^{2}

Also, A + B = (a^{2} – 3b^{2}) + (b^{2} – 3a^{2})

= -2 (a^{2} + b^{2})

If θ is the acute angle between the lines, then

∴ θ = 60°

**Question 5.Find the combined equation of lines passing through the origin each of which making an angle of 30° with the line 3x + 2y – 11 = 0Solution:**

The slope of the line 3x + 2y – 11 = 0 is m

_{1}=

Let m be the slope of one of the lines making an angle of 30° with the line 3x + 2y – 11 = 0.

The angle between the lines having slopes m and m1 is 30°.

On squaring both sides, we get,

∴ (2 – 3m)

^{2}= 3 (2m + 3)

^{2}

∴ 4 – 12m + 9m

^{2}= 3(4m

^{2}+ 12m + 9)

∴ 4 – 12m + 9m

^{2}= 12m

^{2}+ 36m + 27

3m

^{2}+ 48m + 23 = 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

**Question 6.If the angle between lines represented by ax ^{2} + 2hxy + by^{2} = 0 is equal to the angle between lines represented by 2x^{2} – 5xy + 3y^{2} = 0 then show that 100(h^{2} – ab) = (a + b)^{2}.Solution:**

The acute angle θ between the lines ax

^{2}+ 2hxy + by

^{2}= 0 is given by

Comparing the equation 2x

^{2}– 5xy + 3y

^{2}= 0 with ax

^{2}+ 2hxy + by

^{2}= 0, we get,

Let ∝ be the acute angle between the lines 2x

^{2}– 5xy + 3y

^{2}= 0.

This is the required condition.

**Question 7.Find the combined equation of lines passing through the origin and each of which making angle 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.