# Pair of Straight Lines

## Short Questions

If the lines represented by the equation 16x^{2} - kxy + 9y^{2} = 0 are coincident, find the value of k.

Given equation: 16x

^{2}- kxy + 9y

^{2}= 0 --- (i)

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

a = 16, b = 9, h = $\frac{-k}{2}$

Since, lines represented by (i) are coincident,
h^{2} = ab
or, $(\frac{-k}{2})^2$ = 16 × 9

or, k^{2} = 144 × 4

or, k = $\sqrt{576}$

so, k = ±24

Hence, value of `k` is ±24.

If the lines represented by the equation kx^{2} - 24xy + 16y^{2} = 0 are coincident, find the value of k.

Given equation: kx

^{2}- 24xy + 16y

^{2}= 0 --- (i)

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

a = k, b = 16, h = -12

Since, lines represented by (i) are coincident,
h^{2} = ab

or, (-12)^{2} = k × 16

or, $\frac{144}{16}$ = k

so, k = 9

`k`is 9.

If the lines represented by the equation 2x^{2} + 8xy- (m + 1)y^{2}
= 0 are coincident, find the value of 'm'.

Given equation: 2x

^{2}+ 8xy- (m + 1)y

^{2}= 0 --- (i)

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

a = 2, b = -(m + 1), h = 4

Since, lines represented by (i) are coincident,
h^{2} = ab

or, 4^{2} = -2(m + 1)

or, 16 = -2m - 2

or, 18 = -2m

so, m = -9

If the pair of lines represented by the equation x^{2} + 2kxy + 4y^{2} = 0 are coincident, find the value of k.

Given equation: x

^{2}+ 2kxy + 4y

^{2}= 0 --- (i)

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

a = 1, b = 4, h = k

Since, lines represented by (i) are coincident,
h^{2} = ab

or, k^{2} = 1 × 4

or, k = $\sqrt4$

so, k = ±2

Hence, value of 'm' is ±2.

Find the value of 'k' if the pair of lines represented by 3x^{2} - 6xy +
(k + 4)y^{2} = 0 are coincident to each other.

Given equation: 2x

^{2}+ 8xy- (m + 1)y

^{2}= 0 --- (i)

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

a = 3, b = k + 4, h = -3

Since, lines represented by (i) are coincident,
h^{2} = ab

or, (-3)^{2} =3 (k + 4)

or, 9 = 3k + 12

so, k = -1

Find the single equation which represents a pair of straight lines x = 3y and 3x = y.

Given equations:

x - 3y = 0 --- (i)

3x - y = 0 --- (ii)

Now, multiplying (i) and (ii), we get,
(x - 3y)(3x - y) = 0
or, 3x^{2} - xy - 9xy + 3y^{2} = 0

so, 3x^{2} - 10xy + 3y^{2} = 0

Hence, 3x^{2} - 10xy + 3y^{2} = 0 is the required single equation.

Find the separate equations of the pair of lines represented by equation x^{2} + 2x + 2y - y^{2} = 0.

x^{2} + 2x + 2y - y^{2} = 0

or, x^{2} + 2x + 2y - y^{2} = 0

or, x^{2} - y^{2} + 2(x + y) = 0

or, (x + y)(x - y) + 2(x + y) = 0

or, (x + y)(x - y + 2) = 0

Either, x + y = 0 --- (i)

OR, x - y + 2 = 0 --- (ii)

Prove that the angle between a pair of lines represented by 5x^{2} -
6xy - 5y^{2} = 0 is a right angle.

Given equation: 5x

^{2}- 6xy - 5y

^{2}= 0 --- (i)

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

a = 5, b = -5, h = -3

Now, θ be the angle between lines represented by equation (i), so,
tanθ = $\frac{\pm 2\sqrt{h^2\text{ - ab}}}{\text{a + b}}$
or, tanθ = $\frac{\pm 2\sqrt{h^2\text{ - ab}}}{\text{5 - 5}}$

or, tanθ = tan90°

so, θ = 90°

Hence, angle between a pair of lines represented by given equation is a right angle.

Prove that the pair of lines represented by the equation 16x^{2} -
24xy + 9y^{2} = 0 are coincident.

Given equation: 16x

^{2}- 24xy + 9y

^{2}= 0 --- (i)

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

a = 16, h = -12, b = 9

Now, for coincident condition,
h^{2} = ab

or, (-12)^{2} = 16 × 9

so, 144 = 144

Hence, pair of lines represented by given equation is coincident.

Find the separate equations of the pair lines represented by the equation
x^{2} - 2x - y^{2} + 2y = 0.

x

^{2}- 2x - y

^{2}+ 2y = 0 or, x

^{2}- y

^{2}- 2x + 2y = 0

or, (x + y)(x - y) - 2(x - y) = 0

or, (x - y)(x + y - 2) = 0

Either, x - y = 0 --- (i)

OR, x + y - 2 = 0 --- (ii)

Hence, equation (i) and (ii) are separate equation of pair of lines represented by given equation.

Find the value of m, if the lines represented by equation (m - 6)x^{2}
- 8xy + my^{2} = 0 are coincident.

Given equation: (m - 6)x

^{2}- 8xy + my

^{2}= 0 --- (i)

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

a = m - 6, h = -4, b = m

Since, lines represented by (i) are coincident,
h^{2} = ab

or, (-4)^{2} = (m - 6)m

or, 16 = m^{2} - 6m

or, m^{2} - 6m - 16 = 0

or, m^{2} - (8 - 2)m - 16 = 0

or, m^{2} - 8m + 2m - 16 = 0

or, m(m - 8) + 2 (m - 8) = 0

or, (m - 8)(m + 2) = 0

Either, m = 8

OR, m = -2

Hence, value of 'm' is 8 or -2.

If a pair of straight lines represented by the equation kx^{2} - 12xy
+ 3y^{2} = 0 are coincident, find the value of k.

Given equation: kx

^{2}- 12xy + 3y

^{2}= 0 --- (i)

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

a = k, h = -6, b = 3

Since, lines represented by (i) are coincident,
h^{2} = ab

or, (-6)^{2} = k × 3

so, k = 12

Hence, value of 'k' is 12.

If the pair of lines represented by an equation mx^{2} - 5xy - 6y^{2}
= 0 are perpendicular to each other, find the value of m.

Given equation: mx

^{2}- 5xy - 6y

^{2}= 0 --- (i)

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

a = m, h = -$\frac 52$, b = -6

Since, lines represented by (i) are perpendicular,
a + b = 0

or, m - 6 = 0

so, m = 6

Hence, value of 'm' is 6.

Find the angle between the pair of lines represented by equation, 3x^{2}
+ 7xy + 2y^{2} = 0.

Given equation: 3x

^{2}+ 7xy + 2y

^{2}= 0 --- (i)

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

a = 3, h = $\frac 72$, b = 2

Now, θ be the angle between lines represented by equation (i), so,
tanθ = $\frac{\pm 2\sqrt{h^2\text{ - ab}}}{\text{a + b}}$
or, tanθ = $\frac{\pm 2\sqrt{(\frac72)^2\text{ - 6}}}{\text{3 + 2}}$

or, tanθ = $\frac{\pm 2\sqrt{49\text{ - 24}}}{2 \times 5}$

or, tanθ = ±$\frac55$

or, tanθ = ±1

Taking positive,

or, tanθ = tan45°

so, θ = 45°

Taking negative,

or, tanθ = tan135°

so, θ = 135°

Hence, angle between a pair of lines represented by given equation is 45° or 135°.