Long Questions

Here,
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
or, k = ±7
so, positive value of k = 7

Using this value in equation (i), 2x² + 7xy + 3y² = 0 or, 2x² + 6xy + xy + 3y² = 0
or, 2x(x + 3y) + y(x + 3y) = 0
or, (2x + y)(x + 3y) = 0

Hence, the separate equations are 2x + y = 0 and x + 3y = 0.

Here,
Equation of any line perpendicular to the line 2x + 3y = 5 is,
3x - 2y + k = 0 --- (i)

Given straight lines, 3x + 4y = 7 or, x = $\frac{\text{7 - 4y}}{3}$ --- (a)
and, 5x - 2y = 3 --- (b)

Now, substituting the value of 'x' from (a) to (b), $\frac{\text{5(7 - 4y)}}{3}$ - 2y = 3 or, 35 - 20y - 6y = 9
so, y = 1
And, from (a), x = 1

Required equation passes though (1, 1). From (i),
3(1) - 2(1) + k = 0 so, k = -1 --- (ii)

Finally, from (i), 3x - 2y + k = 0 ∴ 3x - 2y - 1 = 0 is the required equation.

Another method

Here, Given straight lines, 3x + 4y = 7 or, x = $\frac{\text{7 - 4y}}{3}$ --- (i)
and, 5x - 2y = 3 --- (ii)

Now, substituting the value of 'x' from (i) to (ii), $\frac{\text{5(7 - 4y)}}{3}$ - 2y = 3 or, 35 - 20y - 6y = 9
so, y = 1
And, from (i), x = 1

Required equation of straight line pass through (1, 1) = (x₁, y₁), so we have, y - y₁ = m₁(x - x₁) or, y - 1 = m₁(x - 1) --- (iii)

Again, slope of line 2x + 3y = 5 is,
m₂ = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = -$\frac 23$

Since, the lines are perpendicular, m₁ × m₂ = -1 or, m₁ × -$\frac 23$ = -1
so, m₁ = $\frac 32$

Finally, from (iii), y - 1 = $\frac 32$(x - 1) or, 2y - 2 = 3x - 3
or, 3x - 2y - 1 = 0

∴ 3x - 2y - 1 = 0 is the required equation of straight line.

 

Let CD be the required equation which is perpendicular bisector of the line segment joining the points (3, 5) and (-7,3).

Now, midpoint of AB = $(\frac{\text{3 + (-7)}}{2}$, $\frac{\text{5 + 3}}{2})$ = (-2 , 4)

So, required equation pass through (-2, 4), y - y₁ = m₁(x - x₁) or, y - 4 = m₁(x + 2) --- (i)

Again, slope of AB (m₂) = $\frac{y_2 \text{ - } y_1}{x_2 \text{ - } x_1}$ = $\frac{\text{3 - 5}}{\text{-7 - 3}}$ = $\frac 15$

Since, the lines are perpendicular, m₁ × m₂ = -1
or, m₁ × $\frac 15$ = -1
so, m₁ = -5

Finally, from (i), y - 4 = -5(x + 2) or, 5x + y - 4 + 10 = 0

∴ 5x + y - 4 + 10 = 0 is the required equation.

Here,
Equation of any line perpendicular to the line 7x - 5y - 6 = 0 is,
5x + 7y + k = 0 --- (i)

It passes though (-1, -2), so, from (i), 5(-1) + 7(-2) + k = 0 or, -5 + 14 + 5 = 0
so, k = 19 --- (ii)

Now, from (i), 5x + 7y + k = 0

∴ 5x + 7y + 19 = 0 is the required equation.

Another method

Here,
Required equation of line pass through (-1, -2). so, y - y₁ = m₁(x - x₁)
or, y + 2 = m₁(x + 1) --- (i)

Now, slope of straight line 7x - 5y - 6 = 0 is,
m₂ = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac 75$

Since, lines are perpendicular, we know, m₁ × m₂ = -1 or, m₁ × $\frac 75$ = -1
so, m₁ = -$\frac 57$

Finally, from (i), y + 2 = -$\frac 57$(x + 1) or, 7y + 14 = -5x - 5
so, 5x + 7y + 19 = 0

Hence, 5x + 7y + 19 = 0 is the required equation.

 

Here,
Equation of any line parallel to the line 2x + y - 4 = 0 is,
2x + y + k = 0 --- (i)

Now, equation makes an intercept of length 2 unit along y-axis. So, it pass through (0, 2). From (i), 2(0) + 2 + k = 0 so, k = -2 --- (ii)

Now, from (i), 2x + y + k = 0 ∴ 2x + y - 2 = 0 is the required equation.

Another method

Here, required equation makes an intercept of length 2 units along y-axis, so, it pass through (0, 2), y - y₁ = m₁(x - x₁)
or, y - 2 = m₁(x - 0) --- (i)

Again, slope of 2x + y - 4 = 0 is,
m₂ = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = -2

Since, lines are parallel,
m₁ = m₂ = 2

Finally, from (i), y - 2 = -2(x) so, 2x + y - 2 = 0

Hence, required equation is 2x + y - 2 = 0.

Here,
Equation of any line parallel to the line x - 2y - 4 = 0 is,
x - 2y + k = 0 --- (i)

It pass through (3, 2), so, from (i), 3 - 2(2) + k = 0 so, k = 1 --- (ii)

Again, from (i),
x - 2y + k = 0 ∴ x - 2y + 1 = 0 is the required equation.

Another method

Here,
Required equation of line pass through (3, 2). So, y - y₁ = m₁(x - x₁)
or, y - 2 = m₁(x - 3) --- (i)

Now, slope of straight line x - 2y - 4 = 0 is,
m₂ = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac 12$

Since, lines are parallel, we know,
m₁ = m₂ = $\frac 12$

Finally, from (i), y - 2 = $\frac 12$(x - 3) or, 2y - 4 = x - 3
so, x - 2y + 1 = 0

Hence, x - 2y + 1 = 0 is the required equation.

 

Here, PQRS is a rhombus.
Slope of diagonal QS (5x

• 7y + 12 = 0) is,
  m₁ = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac 57$

Since, QS is perpendicular to PR, Slope of QS (m₁) × Slope of PR (m₂) = -1
or, $\frac 57$ × m₂ = -1
so, m₂ = -$\frac 75$

Finally, required equation of PR pass through P(2, -3). so, y - y₁ = m₂(x - x₁) or, y + 3 = -$\frac 75$(x - 2)
or, 5y + 12 = -7x + 14
so, 7x + 5y + 1 = 0

Hence, 7x + 5y + 1 = 0 is the required equation.

 

Here, ABCD is a square.
Slope of diagonal AC (3x - 4y + 10 = 0) is,
m₁ = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac 34$

Since, AC is perpendicular to BD, Slope of AC (m₁) × Slope of BD (m₂) = -1 or, $\frac 34$ × m₂ = -1
so, m₂ = -$\frac 43$

Finally, required equation of diagonal BD pass through B(4, -5). so, y - y₁ = m₂(x - x₁) or, y + 5 = -$\frac 43$(x - 4)
or, 3y + 15 = -4x + 16
so, 4x + 3y - 1 = 0

Hence, 4x + 3y - 1 = 0 is the required equation.

  Let CD be the required equation which is perpendicular bisector of the line segment joining the points A(3, -7) and B(-5, 3).

Now, midpoint of AB = $(\frac{\text{3 + (-5)}}{2}$, $\frac{\text{-7 + 3}}{2})$ = (-1 , -2)

So, required equation pass through (-1 , -2), y - y₁ = m₁(x - x₁) or, y + 2 = m₁(x + 1) --- (i)

Again, slope of AB (m₂) = $\frac{y_2 \text{ - } y_1}{x_2 \text{ - } x_1}$ = $\frac{\text{3 - (-7)}}{\text{-5 - 3}}$ = -$\frac 54$

Since, the lines are perpendicular, m₁ × m₂ = -1 or, m₁ × -$\frac 54$ = -1
so, m₁ = $\frac 45$

Finally, from (i), y + 2 = $\frac 45$(x + 1)
or, 5y + 10 = 4x + 4
so, 4x - 5y - 6 = 0

∴ 4x - 5y - 6 = 0 is the required equation.

  Here, A(2, 3) = (x₁, y₁) and B(-4, 1) = (x₂, y₂).
'C' divides the line segment AB in the ratio 2 : 1 (m₁ : m₂).

Now, point 'C' by using section formula is given by,
= $(\frac{m_{1}x_{2} \text{ + } m_{2}x_{1}}{m_1{ + }m_2}$, $\frac{m_{1}y_{2} \text{ + } m_{2}y_{1}}{m_1{ + }m_2})$
= $(\frac{\text{2(-4) + 1(2)}}{\text{2 + 1}}$, $\frac{\text{2(1) + 1(3)}}{\text{2 + 1}})$
= $(\frac {\text{-8 + 2}}{3}$, $\frac 53)$
= (-2, $\frac 53$)

Required equation CD pass through (-2, $\frac 53$) = (x^', y^'), y - y^' = m₁(x - x^')
or, y - $\frac 53$ = m₁(x + 2) --- (i)

Again, slope of line AB,
m₂ = $\frac{y_2 \text{ - } y_1}{x_2 \text{ - } x_1}$ = $\frac{\text{1 - 3}}{\text{-4 - 2}}$ = $\frac13$

Since, AB is perpendicular to CD, m₁ × m₂ = -1
so, m₁ = -3

Finally, from (i), $\frac{\text{3y - 5}}{3}$ = -3(x + 2)
or, 3y - 5 = -9x - 18
so, 9x + 3y + 13 = 0

Hence, 9x + 3y + 13 = 0 is the required equation.

  Here, In triangle PQR,
P(3, 3) = (x₁, y₁), Q(-2, -6) = (x₂, y₂) and R(5, -3) = (x₃, y₃)

Now, centroid of triangle PQR is (C),
= $(\frac{x_1{ + }x_2{ + }x_3}{3}$, $\frac{y_1{ + }y_2{ + }y_3}{3})$
= $(\frac{\text{3 - 2 + 5}}{3}$, $\frac{\text{3 - 6 - 3}}{3})$
= (2, -2)

Required equation pass through (2 , -2), y - y₁ = m₁(x - x₁)
or, y + 2 = m₁(x - 2) --- (i)

Again, slope of line QR (m₂) = y - y₁ = m₁(x - x₁) = $\frac{\text{-3 + 6}}{\text{5 + 2}}$ = $\frac 37$

Since, QR is parallel with ST,
m₁ = m₂ = $\frac 37$

Finally, from (i), y + 2 = $\frac37$(x - 2) or, 7y + 14 = 3x - 6
so, 3x - 7y - 20 = 0

Hence, 3x - 7y - 20 = 0 is the required equation.

Given,
In triangle ABC, A(3, 2) = (x₁, y₁), B(1, -1) = (x₂, y₂) and C(5, -5) = (x₃, y₃)

Now, for centroid of triangle ABC (C),
= $(\frac{x_1{ + }x_2{ + }x_3}{3}$, $\frac{y_1{ + }y_2{ + }y_3}{3})$
= $(\frac{\text{3 + 1 + 5}}{3}$, $\frac{\text{2 - 1 - 5}}{3})$
= (3, -$\frac43$)

So, required equation pass through (3, -$\frac43$),
y - y₁ = m₁(x - x₁) or, y + $\frac43$ = m₁(x - 3) --- (i)

Again, slope of line BC (m₂) = y - y₁ = m₁(x - x₁) = $\frac{\text{-5 + 1}}{\text{5 - 1}}$ = -1

Since, BC is parallel with ST,
m₁ = m₂ = -1

Finally, from (i), y + $\frac43$ = -1(x - 3)
or, 3y + 4 = -3x + 9
so, 3x + 4y - 5 = 0

Hence, 3x + 4y - 5 = 0 is the required equation.

Let CD be the required equation which is perpendicular bisector of the line segment joining the points A(4, -5) and B(-8, 9).

Now, midpoint of AB = $(\frac{\text{4 + (-8)}}{2}$, $\frac{\text{-5 + 9}}{2})$ = (-2 , 2)

Required equation pass through (-1 , -2), y - y₁ = m₁(x - x₁) or, y - 2 = m₁(x + 2) --- (i)

Again, slope of AB (m₂) = $\frac{y_2 \text{ - } y_1}{x_2 \text{ - } x_1}$ = $\frac{\text{9 - (-5)}}{\text{-8 - 4}}$ = -$\frac 76$

Since, the lines are perpendicular, m₁ × m₂ = -1 or, m₁ × -$\frac 76$ = -1
so, m₁ = $\frac 67$

Finally, from (i), y - 2 = $\frac 67$(x + 2) or, 7y - 14 = 6x + 12
so, 6x - 7y + 26 = 0

∴ 6x - 7y + 26 = 0 is the required equation.

Here,
Required equation of line pass through (-6, 2). So, y - y₁ = m₁(x - x₁)
or, y - 2 = m₁(x + 6) --- (i)

Now, slope of line joining the points (-3, 2) and (5, -3) is,
m₂ = $\frac{y_2 \text{ - } y_1}{x_2 \text{ - } x_1}$ = $\frac{\text{-3 - 2}}{\text{5 + 3}}$ = -$\frac 58$

Since, lines are perpendicular, we know, m₁ × m₂ = -1 or, m₁ × -$\frac 58$ = -1
so, m₁ = $\frac85$

Finally, from (i), y - 2 = $\frac85$(x + 6) or, 5y - 10 = 8x + 48
so, 8x - 5y + 58 = 0

Hence, 8x - 5y + 58 = 0 is the required equation.

Here,
Equation of any line parallel to the line 3x - 2y = 5 is,
3x - 2y + k = 0 --- (i)

Now, midpoint of the line segment joining the points (-4, 2) and (2, 4),
(x₁, y₁) = $(\frac{\text{-4 + 2}}{2}$, $\frac{\text{2 + 4}}{2})$ = (-1, 3)

Required equation pass through (-1, 3). From (i), 3(-1) - 2(3) + k = 0 or, -3 - 6 + k = 0
so, k = 9 --- (ii)

Now, from (i),
3x - 2y + k = 0

∴ 3x - 2y + 9 = 0 is the required equation.

Another method

Here, midpoint of the line segment joining the points (-4, 2) and (2, 4),
(x₁, y₁) = $(\frac{\text{-4 + 2}}{2}$, $\frac{\text{2 + 4}}{2})$ = (-1, 3)

So, required equation pass through (-1, 3), y - y₁ = m₁(x - x₁)
or, y - 3 = m₁(x + 1) --- (i)

Now, slope of line 3x - 2y = 5,
m₁ = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = $\frac32$

Since, lines are parallel,
m₁ = m₂ = $\frac32$

Finally, from (i), y - 3 = $\frac32$(x + 1) or, 2y - 6 = 3x + 3
so, 3x - 2y + 9 = 0

Hence, 3x - 2y + 9 = 0 is the required equation.

 

Here,
Equation of any line parallel to the line y = x - 6 ⇒ x - y - 6 = 0 is,
x - y + k = 0 --- (i)

Now, solving 'x' and 'y' from (i) and (ii), 2(3 - y) - 3y = 1 or, 6 - 2y - 3y = 1
or, 5 = 5y
so, y = 1 and x = 2 (from i)

So, equation (i) pass through (2, 1), 2 - 1 + k = 0 ∴ k = -1

Now, from (i), x - y - k = 0 or, x - y - 1 = 0
or, x - y = 1
or, $\frac{x}{1}$ + $\frac{y}{-1}$ = 1 --- (ii)

Here, equation (ii) is equal to the given equation $\frac xa$ + $\frac yb$ = 1, so, comparing the values, we get,
a = 1 and b = -1

Hence, value of a = 1 and b = -1.

Another method

Here, given equation of lines, x + y = 3 or, x = 3 - y --- (i) and, 2x - 3y = 1 --- (ii)

Now, solving 'x' and 'y' from (i) and (ii), 2(3 - y) - 3y = 1 or, 6 - 2y - 3y = 1
or, 5 = 5y
so, y = 1 and x = 2 (from i)

Any line that pass through (x, y) = (2, 1) is, y - y₁ = m₁(x - x₁)
or, y - 1 = m₁(x - 2) --- (iii)

Again, slope of line y = x - 6 $\Rightarrow$ x - y - 6 = 0 --- (iv) is,
m₂ = -$\frac{\text{Coefficient of x}}{\text{Coefficient of y}}$ = 1

Since (iii) and (iv) are parallel,
m₁ = m₂ = 1

From (iii), y - 1 = (1)x - 2 or, x - y = 1
or, $\frac{x}{1}$ + $\frac{y}{-1}$ = 1 --- (v)

Here, equation (ii) is equal to the given equation $\frac xa$ + $\frac yb$ = 1, so, comparing the values, we get,
a = 1 and b = -1

Hence, value of a = 1 and b = -1.