# Long Questions

Find the equation of any one straight line passing through the point (4, -1) and making an angle of 45° with the line 2x - 3y = 5.

Here,
The required straight line pass through the point (4, -1), so, we have, (y - y1) = m1(x - x1) or, y + 1 = m1(x - 4) --- (i)

Also, the required line makes an angle of θ = 45° with the line 2x - 3y = 5 whose slope,
m2 = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac{2}{3}$.

So, the angles between these lines is given by, tanθ = ±$\frac{m_1\text{ - }m_2}{1 \text{ + }m_1 m_2}$ or, tan45° = ±$\frac{m_1\text{ - }\frac{2}{3}}{1 \text{ + }m_1 \frac{2}{3}}$
or, 1 = ±$\frac{3m_1\text{ - 2}}{3 \text{ + }2m_1}$
or, 1 = ±$\frac{3m_1\text{ - 2}}{3 \text{ + }2m_1}$
or, 3 + 2m1 = ±3m1 - 2

Taking positive sign,
3 + 2m1 = 3m1 - 2
so, m1 = 5
Using its value at (i),
y + 1 = 5(x - 4)
so, 5x - y - 21 = 0 --- (A)

Taking negative sign,
3 + 2m1 = -(3m1 - 2)
or, 5m1 = -1
so, m1 = -$\frac{1}{5}$
Using its value at (i),
y + 1 = -$\frac{1}{5}$(x - 4)
or, 5y + 5 = -x + 4
so, x + 5y + 1 = 0 --- (B)

Hence, from (A) and (B), 5x - y - 21 = 0 and x + 5y + 1 = 0 are required equations.

If the acute angle between the pair of straight lines represented by x2 - 2xycosecθ + y2 = 0 is α, then prove that: α = 90° - θ.

Here,
Given equation: x2 - 2xycosecθ + y2 = 0 --- (i)

Comparing it with ax2 + 2hxy + by2 = 0, we get,
a = 1, h = -cosecθ, b = 1

Now, α be the acute angle between lines represented by equation (i), we have, tanα = +$\frac{2\sqrt{h^2\text{ - ab}}}{\text{a + b}}$ or, tanα = +$\frac{2\sqrt{(-cosec^2\theta) \text{ - 1}}}{\text{1 + 1}}$
or, tanα = $\sqrt{cosec^2\theta \text{ - 1}}$
or, tanα = $\sqrt{cot^2\theta}$
or, tanα = cotθ
or, tanα = tan(90° - θ)
so, α = 90° - θ

Hence, it is proved that the acute angle between a pair of lines represented by the given equation is 90° - θ.

Note: Acute angle is an angle less than 90°. So, we take only positive sign. If there was obtuse angle in the question then we would take negative sign.

Find the angles between the lines represented by the equation x2 + 4xy + y2 = 0.

Here,
Given equation: x2 + 4xy + y2 = 0 --- (i)

Comparing it with ax2 + 2hxy + by2 = 0, we get,
a = 1, h = 2, b = 1

Let, θ be the angle between lines represented by equation (i), we know, tanθ = $\frac{\pm 2\sqrt{h^2\text{ - ab}}}{\text{a + b}}$ or, tanθ = $\frac{\pm 2\sqrt{2^2\text{ - 1}}}{2}$
or, tanθ = ±$\sqrt3$
Taking positive,
or, tanθ = tan60°
so, θ = 60°
Taking negative,
or, tanθ = -tan60°
or, tanθ = tan(180° - 60°)
so, θ = 120°

Hence, the angle between a pair of lines represented by the given equation is 60° or 120°.

Find the separate equations of the pair of lines represented by the equation 6x2 - xy - y2 = 0. Also find the angle between them.

Given equation: 6x2 - xy - y2 = 0 --- (i)

First part: 6x2 - xy - y2 = 0
or, 6x2 - (3 - 2)xy - y2 = 0
or, 6x2 - 3xy + 2xy - y2 = 0
or, 3x(2x - y) + y(2x - y) = 0
or, (2x - y)(3x + y) = 0
Either, 2x - y = 0
OR, 3x + y = 0

So, the separate equations of the pair of lines represented by (i) are 2x - y = 0 and 3x + y = 0.

Second part:
Comparing equation (i) with ax2 + 2hxy + by2 = 0, we get,
a = 6, h = -$\frac12$, b = -1

Let θ be the angle between lines represented by equation (i), we have, tanθ = $\frac{\pm 2\sqrt{h^2\text{ - ab}}}{\text{a + b}}$ or, tanθ = $\frac{\pm 2\sqrt{(-\frac12)^2\text{ - (-6)}}}{\text{6 - 1}}$
or, tanθ = $\frac{\pm 2\sqrt{\text{1 + 24}}}{2 \times 5}$
or, tanθ = $\frac{\pm \sqrt{25}}{5}$
or, tanθ = ±1

Taking positive,
or, tanθ = tan45°
so, θ = 45°

Taking negative,
or, tanθ = -tan45°
or, tanθ = tan(180° - 45°)
so, θ = 135°

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

The angle between a pair of straight lines is represented by the equation x2 + xy - ky2 = 0 is 45°. Find the value of k.

Here,
Given equation: x2 + xy - ky2 = 0 --- (i)

Comparing it with ax2 + 2hxy + by2 = 0, we get,
a = 1, h = $\frac 12$, b = -k

Since 45° is the angle between lines represented by equation (i), we have, tanθ = $\frac{\pm 2\sqrt{h^2\text{ - ab}}}{\text{a + b}}$ or, tan45° = $\frac{\pm 2\sqrt{(\frac12)^2\text{ - (-k)}}}{\text{1 - k}}$
or, 1 = $\frac{\pm 2\sqrt{\text{1 + 4k}}}{\text{2(1 - k)}}$
or, 1(1 - k) = ±$\sqrt{\text{1 + 4k}}$
Squaring on both sides, we get,
or, 1 - 2k + k2 = 1 + 4k
or, k2 - 6k = 0
or, k(k - 6) = 0
Either, k = 0
OR, k = 6

Hence, the value of k is 0 or 6.

Find the equations of the pair of lines represented by the equation 2x2 + 3xy + y2 = 0 and find the angle between them.

Given equation: 2x2 + 3xy + y2 = 0 --- (i)

First part: 2x2 + 3xy + y2 = 0
or, 2x2 + (2 + 1)xy + y2 = 0
or, 2x2 + 2xy + xy + y2 = 0
or, 2x(x + y) + y(x + y) = 0
or, (x + y)(2x + y) = 0
Either, 2x + y = 0
OR, x + y = 0

So, the separate equations of the pair of lines represented by (i) are 2x + y = 0 and x + y = 0.

Second part:
Comparing equation (i) with ax2 + 2hxy + by2 = 0, we get,
a = 2, h = $\frac32$, b = 1

Let θ be the angle between lines represented by equation (i), we have, tanθ = $\frac{\pm 2\sqrt{h^2\text{ - ab}}}{\text{a + b}}$ or, tanθ = $\frac{\pm 2\sqrt{(\frac32)^2\text{ - 2}}}{\text{2 + 1}}$
or, tanθ = $\frac{\pm 2\sqrt{\text{9 - 8}}}{2 \times 3}$
or, tanθ = ±$\frac13$
so, θ = tan-1$(\pm\frac13)$

Hence, angle between a pair of lines represented by given equation is tan-1$(\pm\frac13)$.

If the angle between the lines represented by 2x2 + kxy + 3y2 = 0 is 45°, find the value of k.

Given equation: 2x2 + kxy + 3y2 = 0 --- (i)

Comparing it with ax2 + 2hxy + by2 = 0, we get,
a = 2, h = $\frac k2$, b = 3

Since 45° is the angle between lines represented by equation (i), we have, tanθ = $\frac{\pm 2\sqrt{h^2\text{ - ab}}}{\text{a + b}}$ or, tan45° = $\frac{\pm 2\sqrt{(\frac k2)^2\text{ - 6}}}{\text{2 + 3}}$ or, 1 = $\frac{\pm 2\sqrt{k^2\text{ - 24}}}{2 \times 5}$ or, 5 = ±$\sqrt{k^2\text{ - 24}}$
Squaring on both sides, we get,
or, 25 = k2 - 24
or, 49 = k2
so, k = ±7
Hence, the value of k is ±7.

If the straight lines represented by 5x2 - 8xy + py2 = 0 are perpendicular to each other, then find the value of p. Also find the equations of straight lines represented by 5x2 - 8xy + (p + 8)y2 = 0.

First Part:
Given equation: 5x2 - 8xy + py2 = 0 --- (i)

Comparing it with ax2 + 2hxy + by2 = 0, we get,
a = 5, h = -4, b = p

Since the lines represented by equation (i) are perpendicular, a + b = 0 or, 5 + p = 0
so, p = 5 --- (ii)

Second Part:
Given equation: 5x2 - 8xy + (p + 8)y2 = 0

Using (ii), 5x2 - 8xy + (-5 + 8)y2 = 0 (from ii) or, 5x2 - 8xy + 3y2 = 0
or, 5x2 - (5 + 3)xy + 3y2 = 0
or, 5x2 - 5xy - 3xy + 3y2 = 0
or, 5x(x - y) - 3y(x - y) = 0
or, (x - y)(5x - 3y) = 0
Either, x - y = 0
OR, 5x - 3y = 0

Hence, the separate equations of the pair of lines represented by the given equation are x - y = 0 and 5x - 3y = 0.

Find the equation of lines which pass through the point (1, 2) and perpendicular to the lines represented by x2 - 5xy + 4y2 = 0.

Required equation pass through the point (1, 2), so, y - y1 = m1(x - 1)
or, y - 2 = m1(x - 1) --- (i)

Now, for separate equations, x2 - 5xy + 4y2 = 0 or, x2 - (4 + 1)xy + 4y2 = 0
or, x2 - 4xy - xy + 4y2 = 0
or, x(x - 4y) - y(x - 4y) = 0
or, (x - 4y)(x - y) = 0
Either, x - y = 0 --- (ii)
OR, x - 4y = 0 --- (iii)

From (ii),
slope (m2) = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = 1

Since (i) and (ii) are perpendicular, m1 × m2 = -1
or, m1 = -1

So, from (i), y - 2 = -1(x - 1)
or, y - 2 = -x + 1 so, x + y - 3 = 0 --- (A)

Again, from (iii),
slope (m3) = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac 14$

Since (i) and (iii) are also perpendicular, m1 × m3 = -1 or, m1 = -4

Again, from (i), y - 2 = -4(x - 1)
or, y - 2 = -4x + 4
so, 4x + y - 6 = 0 --- (B)

Finally, multiplying (A) and (B), (x + y - 3)(4x + y - 6) = 0 or, 4x2 + xy - 6x + 4xy + y2 - 6y - 12x - 3y + 18 = 0
or, 4x2 + 5xy - 18x - 9y + 18 + y2 = 0
so, 4x2 + 5xy + y2 - 18x - 9y + 18 = 0

Hence, 4x2 + 5xy + y2 - 18x - 9y + 18 = 0 is the required equation.

Find the single equation of a pair of straight lines passing through the origin and perpendicular to the lines represented by the equation 2x2 - 7xy + 5y2 = 0.

Required equation pass through the point (0, 0), so, y - y1 = m1(x - 1)
or, y - 0 = m1(x - 0)
or, y = m1x --- (i)

Now, for separate equations, 2x2 - 7xy + 5y2 = 0 or, 2x2 - (5 + 2)xy + 5y2 = 0
or, 2x2 - 5xy - 2xy + 5y2 = 0
or, x(2x - 5y) - y(2x - 5y) = 0
or, (2x - 5y)(x - y) = 0
Either, 2x - 5y= 0 --- (ii)
OR, x - y = 0 --- (iii)

From (ii),
slope (m2) = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac 25$

Since (i) and (ii) are perpendicular, m1 × m2 = -1
or, m1 = -$\frac 52$

So, from (i), y = -$\frac 52$x so, 5x + 2y = 0 --- (A)

Again, from (iii),
slope (m3) = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = 1

Since (i) and (iii) are also perpendicular, m1 × m3 = -1
or, m1 = -1

Again, from (i), y = -1x
so, x + y = 0 --- (B)

Finally, multiplying (A) and (B), (5x + 2y)(x + y) = 0 or, 5x2 + 5xy + 2xy + 2y2 = 0
so, 5x2 + 7xy + 2y2 = 0

Hence, 5x2 + 7xy + 2y2 = 0 is the required equation.

Find the single equation of a pair of straight lines passing through the origin and perpendicular to the lines represented by the equation 2x2 - 3xy - 5y2 = 0.

Required equation pass through the point (0, 0), so, y - y1 = m1(x - 1)
or, y - 0 = m1(x - 0) or, y = m1x --- (i)

Now, for separate equations, 2x2 - 3xy - 5y2 = 0 or, 2x2 - (5 - 2)xy - 5y2 = 0
or, 2x2 - 5xy + 2xy - 5y2 = 0
or, x(2x - 5y) + y(2x - 5y) = 0
or, (2x - 5y)(x + y) = 0
Either, 2x - 5y = 0 --- (ii)
OR, x + y = 0 --- (iii)

From (ii),
slope (m2) = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac 25$

Since (i) and (ii) are perpendicular, m1 × m2 = -1
or, m1 = -$\frac 52$

So, from (i), y = -$\frac 52$x so, 5x + 2y = 0 --- (A)

Again, from (iii),
slope (m3) = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = -1

Since (i) and (iii) are also perpendicular, m1 × m3 = -1 or, m1 = 1

Again, from (i), y = x so, x - y = 0 --- (B)

Finally, multiplying (A) and (B), (5x + 2y)(x - y) = 0 or, 5x2 - 5xy + 2xy - 2y2 = 0
so, 5x2 - 3xy - 2y2 = 0

Hence, 5x2 - 3xy - 2y2 = 0 is the required equation.

If the two straight line represented by an equation 3x2 + 8xy + my2 = 0 are perpendicular to each other, find the separate equation of two lines.

Given equation: 3x2 + 8xy + my2 = 0 --- (i)

Comparing it with ax2 + 2hxy + by2 = 0, we get,
a = 3, h = 4, b = m

Since lines represented by equation (i) are perpendicular, a + b = 0
or, 3 + m = 0
so, m = -3 --- (ii)

Now, for separate equations, 3x2 + 8xy + my2 = 0 or, 3x2 + 8xy - 3y2 = 0 (from i)
or, 3x2 + (9 - 1)xy - 3y2 = 0
or, 3x2 + 9xy - xy - 3y2 = 0
or, 3x(x + 3y) - y(x + 3y) = 0
or, (3x - y)(x + 3y) = 0
Either, 3x - y = 0
OR, x + 3y = 0

Hence, the separate equations of the pair of lines represented by (i) are 3x - y = 0 and x + 3y = 0.

Find the single equation of the pair of straight lines passing through the origin and perpendicular to the pair of lines represented by an equation 2x2 - 3xy + y2 = 0.

Required equation pass through the point (0, 0), so, y - y1 = m1(x - 1) or, y - 0 = m1(x - 0) or, y = m1x --- (i)

Now, for separate equations, 2x2 - 3xy + y2 = 0 or, 2x2 - (2 + 1)xy + y2 = 0
or, 2x2 - 2xy - xy + y2 = 0
or, 2x(x - y) - y(x - y) = 0
or, (2x - y)(x - y) = 0
Either, 2x - y = 0 --- (ii)
OR, x - y = 0 --- (iii)

From (ii),
slope (m2) = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = 2

Since (i) and (ii) are perpendicular, m1 × m2 = -1
or, m1 = -$\frac 12$

So, from (i), y = -$\frac 12$x
so, x + 2y = 0 --- (A)

Again, from (iii),
slope (m3) = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = 1

Since (i) and (iii) are also perpendicular, m1 × m3 = -1
or, m1 = -1

Again, from (i), y = -x so, x + y = 0 --- (B)

Finally, multiplying (A) and (B), (x + 2y)(x + y) = 0 or, x2 + xy + 2xy + 2y2 = 0
So, x2 + 3xy + 2y2 = 0

Hence, x2 + 3xy + 2y2 = 0 is the required equation.

Find the acute angle between the pair of lines given by 3x2 + 7xy + 2y2 = 0.

Given equation: 3x2 + 7xy + 2y2 = 0 --- (i)

Comparing it with ax2 + 2hxy + by2 = 0, we get,
a = 3, h = $\frac 72$, b = 2

Let, θ be the acute angle between lines represented by equation (i), we have, tanθ = +$\frac{2\sqrt{h^2\text{ - ab}}}{\text{a + b}}$ or, tanθ = +$\frac{2\sqrt{\text{49 - 24}}}{\text{2(3 + 2)}}$
or, tanθ = +$\frac 55$
or, tanθ = +1
or, tanθ = tan45°
so, θ = 45°

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