Equation of a Straight Line — Exercise 12.2

State which one of the following is true :
The straight lines y = 3x - 5 and 2y = 4x + 7 are

(i) parallel

(ii) perpendicular

(iii) neither parallel nor perpendicular.

Answer

Lines are y = 3x - 5 and 2y = 4x + 7 or y = 2x + $\dfrac{7}{2}$.

Comparing y = 3x - 5 and y = 2x + $\dfrac{7}{2}$ with y = mx + c we get,

slopes = 3 and 2.

Since, slope of both the lines are neither equal nor their products is -1. Thus, the lines are neither parallel nor perpendicular.

∴ Option (iii) is correct

Hence, the straight lines y = 3x - 5 and 2y = 4x + 7 are neither parallel nor perpendicular.

If 6x + 5y - 7 = 0 and 2px + 5y + 1 = 0 are parallel lines, find the value of p.

Answer

Converting 6x + 5y - 7 = 0 in the form y = mx + c we get,

⇒ 6x + 5y - 7 = 0

⇒ 5y = -6x + 7

⇒ y = $-\dfrac{6}{5}x + \dfrac{7}{5}$

Comparing, we get slope of this line = m₁ = $-\dfrac{6}{5}$.

Converting 2px + 5y + 1 = 0 in the form y = mx + c we get,

⇒ 2px + 5y + 1 = 0

⇒ 5y = -2px - 1

⇒ y = $-\dfrac{2p}{5}x - \dfrac{1}{5}$

Comparing, we get slope of this line = m₂ = $-\dfrac{2p}{5}$

Given, two lines are parallel so their slopes will be equal,

m₁ = m₂

$\Rightarrow -\dfrac{6}{5} = -\dfrac{2p}{5} \\[1em] \Rightarrow 2p = 6 \\[1em] \Rightarrow p = 3.$

Hence, the value of p = 3.

If the straight lines 3x - 5y + 7 = 0 and 4x + ay + 9 = 0 are perpendicular to one another, find the value of a.

Answer

Converting 3x - 5y + 7 = 0 in the form y = mx + c we get,

⇒ 3x - 5y + 7 = 0

⇒ 5y = 3x + 7

⇒ y = $\dfrac{3}{5}x + \dfrac{7}{5}$

Comparing, we get slope of first line = m₁ = $\dfrac{3}{5}$.

Converting 4x + ay + 9 = 0 in the form y = mx + c we get,

⇒ 4x + ay + 9 = 0

⇒ ay = -4x - 9

⇒ y = $-\dfrac{4}{a}x - \dfrac{9}{a}$

Comparing, we get slope of second line = m₂ = $-\dfrac{4}{a}$

Given, two lines are perpendicular so product of their slopes will be equal to -1,

m₁.m₂ = -1

$\Rightarrow \dfrac{3}{5} \times -\dfrac{4}{a} = -1 \\[1em] \Rightarrow -\dfrac{12}{5a} = -1 \\[1em] \Rightarrow a = \dfrac{12}{5}.$

Hence, the value of a = $\dfrac{12}{5}$.

If the lines 3x + by + 5 = 0 and ax - 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.

Answer

Given,

Lines 3x + by + 5 = 0 and ax - 5y + 7 = 0 are perpendicular to each other. Then the product of their slopes is -1.

Converting 3x + by + 5 = 0 in the form y = mx + c.

⇒ by = -3x - 5

⇒ y = $-\dfrac{3}{\text{b}}\text{x} - \dfrac{5}{\text{b}}$.

Comparing with y = mx + c we get,

Slope of first line = m₁ = $-\dfrac{3}{\text{b}}$.

Converting ax - 5y + 7 = 0 in the form y = mx + c.

⇒ 5y = ax + 7

⇒ y = $\dfrac{\text{a}}{5}\text{x} + \dfrac{7}{5}$.

Comparing with y = mx + c we get,

Slope of second line = m₂ = $\dfrac{\text{a}}{5}$.

For perpendicular lines, product of their slopes is -1.

∴ m₁.m₂ = -1.

$\Rightarrow -\dfrac{3}{b} \times \dfrac{a}{5} = -1 \\[1em] \Rightarrow -\dfrac{3a}{5b} = -1 \\[1em] \Rightarrow -3a = -5b \\[1em] \Rightarrow 3a = 5b.$

Hence, the relation between a and b is given by 3a = 5b.

Is the line through (-2, 3) and (4, 1) perpendicular to the line 3x = y + 1? Does the line 3x = y + 1 bisect the join of (-2, 3) and (4, 1)?

Answer

Equation of line through (-2, 3) and (4, 1) can be given by two-point form i.e.,

$y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Putting values in above formula we get,

$\Rightarrow y - 3 = \dfrac{1 - 3}{4 - (-2)}(x - (-2)) \\[1em] \Rightarrow y - 3 = \dfrac{-2}{6}(x + 2) \\[1em] \Rightarrow y - 3 = \dfrac{-1}{3}(x + 2) \\[1em] \Rightarrow y - 3 = -\dfrac{1}{3}x - \dfrac{2}{3} \\[1em] \Rightarrow y = -\dfrac{1}{3}x - \dfrac{2}{3} + 3 \\[1em] \Rightarrow y = -\dfrac{1}{3}x - \dfrac{2 + 9}{3} \\[1em] \Rightarrow y = -\dfrac{1}{3}x - \dfrac{11}{3}.$

Comparing the above equation with y = mx + c we get,

slope = m₁ = $-\dfrac{1}{3}$

The other equation is 3x = y + 1 or y = 3x - 1, comparing this with y = mx + c we get,

slope = m₂ = 3.

Product of slopes,

$= m_1 \times m_2 \\[1em] = -\dfrac{1}{3} \times 3 \\[1em] = -1.$

Since, the product of slopes is -1 hence, the lines are perpendicular to each other.

Mid-point of (-2, 3) and (4, 1) can be given by mid-point formula i.e.,

$\Big(\dfrac{-2 + 4}{2}, \dfrac{3 + 1}{2}\Big)$ = (1, 2).

Line 3x = y + 1 bisects the line joining (-2, 3) and (4, 1) if the mid-point i.e., (1, 2) satisfies the equation.

Putting (1, 2) in 3x = y + 1.

L.H.S. = 3x = 3(1) = 3.

R.H.S. = y + 1 = 2 + 1 = 3.

Since, L.H.S. = R.H.S. hence, (1, 2) satisfies 3x = y + 1.

Hence, the line 3x = y + 1 is perpendicular to the line joining (-2, 3) and (4, 1) and also bisects it.

The line through A(-2, 3) and B(4, b) is perpendicular to the line 2x - 4y = 5. Find the value of b.

Answer

Slope of the line = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Slope of line passing through A and B is m₁,

$= \dfrac{b - 3}{4 - (-2)} \\[1em] = \dfrac{b - 3}{6}.$

The equation of other line is,

⇒ 2x - 4y = 5

⇒ 4y = 2x - 5

⇒ y = $\dfrac{2}{4}x - \dfrac{5}{4}$

⇒ y = $\dfrac{1}{2}x - \dfrac{5}{4}$

Comparing the equation with y = mx + c we get,

slope = m₂ = $\dfrac{1}{2}$

As lines are perpendicular to each other, we have

$\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow \dfrac{b - 3}{6} \times \dfrac{1}{2} = -1 \\[1em] \Rightarrow \dfrac{b - 3}{12} = -1 \\[1em] \Rightarrow b - 3 = -12 \\[1em] \Rightarrow b = -12 + 3 \\[1em] \Rightarrow b = -9.$

Hence, the value of b is -9.

If the lines 3x + y = 4, x - ay + 7 = 0 and bx + 2y + 5 = 0 form three consecutive sides of a rectangle, find the values of a and b.

Answer

Given lines are :

3x + y = 4 ....(i)

x- ay + 7 = 0 ....(ii)

bx + 2y + 5 = 0 ....(iii)

It's said that these lines form three consecutive sides of a rectangle.

So,

Lines (i) and (ii) must be perpendicular and also (ii) and (iii) will be perpendicular.

Slope of line (i) is

⇒ 3x + y = 4

⇒ y = -3x + 4.

Comparing with y = mx + c we get,

slope = m₁ = -3.

Slope of line (ii) is

⇒ x - ay + 7 = 0

⇒ ay = x + 7

⇒ y = $\dfrac{1}{a}x + \dfrac{7}{a}$

Comparing with y = mx + c we get,

slope = m₂ = $\dfrac{1}{a}$.

Slope of line (iii) is

⇒ bx + 2y + 5 = 0

⇒ 2y = -bx - 5

⇒ y = $-\dfrac{b}{2}x - \dfrac{5}{2}$

Comparing with y = mx + c we get,

slope = m₃ = $-\dfrac{b}{2}$.

Since, lines (i) and (ii) are perpendicular so,

⇒ m₁ × m₂ = -1

$\Rightarrow -3 \times \dfrac{1}{a} = -1 \\[1em] \Rightarrow a = \dfrac{-3}{-1} \\[1em] \Rightarrow a = 3.$

Since, lines (ii) and (iii) are perpendicular so,

⇒ m₂ × m₃ = -1

$\Rightarrow \dfrac{1}{a} \times -\dfrac{b}{2} = -1 \\[1em] \Rightarrow -\dfrac{b}{2a} = -1 \\[1em] \Rightarrow b = 2a \\[1em] \Rightarrow b = 2(3) \\[1em] \Rightarrow b = 6.$

Thus, the value of a is 3 and the value of b is 6.

Find the value of 'p' if the lines 5x - 3y + 2 = 0 and 6x - py + 7 = 0 are perpendicular to each other. Hence find the equation of a line passing through (-2, -1) and parallel to 6x - py + 7 = 0.

Answer

Given lines,

⇒ 5x - 3y + 2 = 0 and 6x - py + 7 = 0

⇒ 3y = 5x + 2 and py = 6x + 7

⇒ y = $\dfrac{5}{3}x + \dfrac{2}{3} \text{ and } y = \dfrac{6}{p}x + \dfrac{7}{p}$

Comparing above equations with y = mx + c we get,

Slope of 1st line = $\dfrac{5}{3}$

Slope of 2nd line = $\dfrac{6}{p}$

Since, product of slopes of perpendicular lines = -1.

$\therefore \dfrac{5}{3} \times \dfrac{6}{p} = -1 \\[1em] \Rightarrow 5 \times \dfrac{2}{p} = -1 \\[1em] \Rightarrow \dfrac{10}{p} = -1 \\[1em] \Rightarrow p = -10 \\[1em]$

Given,

⇒ 6x - py + 7 = 0

⇒ 6x - (-10)y + 7 = 0

⇒ 6x + 10y + 7 = 0

⇒ 10y = -6x - 7

⇒ y = $-\dfrac{6}{10}x - \dfrac{7}{10}$

Comparing above equations with y = mx + c we get,

Slope = $-\dfrac{6}{10}$

Since, parallel lines have equal slope.

∴ Slope of line parallel to line 6x + 10y + 7 = 0 is $-\dfrac{6}{10}$

By point-slope form,

⇒ y - y₁ = m(x - x₁)

Equation of line passing through (-2, -1) and slope $-\dfrac{6}{10}$ is

$\Rightarrow y - (-1) = -\dfrac{6}{10}[x - (-2)] \\[1em] \Rightarrow 10(y + 1) = -6(x + 2) \\[1em] \Rightarrow 10y + 10 = -6x - 12 \\[1em] \Rightarrow 10y + 10 + 6x + 12 = 0 \\[1em] \Rightarrow 6x + 10y + 22 = 0 \\[1em] \Rightarrow 2(3x + 5y + 11) = 0 \\[1em] \Rightarrow 3x + 5y + 11 = 0$

Hence, p = -10 and the equation of line is 3x + 5y + 11 = 0.

Find the equation of a line, which has the y-intercept 4, and is parallel to the line 2x - 3y - 7 = 0. Find the coordinates of the point where it cuts the x-axis.

Answer

Given equation of line,

⇒ 2x - 3y - 7 = 0,

Converting it in the form y = mx + c,

⇒ 3y = 2x - 7

⇒ y = $\dfrac{2}{3}x - \dfrac{7}{3}$.

Comparing with y = mx + c, slope = $\dfrac{2}{3}$.

Since,the other line is parallel so, its slope will also be equal to $\dfrac{2}{3}$. Given, y-intercept is 4 or c = 4.

Putting values of slope and y-intercept in y = mx + c, we will get the equation of line as,

⇒ y = $\dfrac{2}{3}x + 4$

⇒ y = $\dfrac{2x + 12}{3}$

⇒ 3y = 2x + 12

⇒ 2x - 3y + 12 = 0.

At the point where the line intersects the x-axis, the y-coordinate there will be zero. So, putting y = 0 in 2x - 3y + 12 = 0.

⇒ 2x - 3(0) + 12 = 0
⇒ 2x = -12
⇒ x = -6.

∴ Coordinates = (-6, 0).

Hence, the equation of the line is 2x - 3y + 12 = 0 and it intersects the x-axis at (-6, 0).

Find the equation of a straight line perpendicular to the line 2x + 5y + 7 = 0 and with y-intercept -3.

Answer

Given equation of line,

⇒ 2x + 5y + 7 = 0

Converting it in the form y = mx + c,

⇒ 5y = -2x - 7

⇒ y = $-\dfrac{2}{5}x - \dfrac{7}{5}$.

Comparing with y = mx + c we get,

m = $-\dfrac{2}{5}$

Let slope of other line be m', since lines are perpendicular so,

⇒ m × m' = -1

$\Rightarrow -\dfrac{2}{5} \times m' = -1 \\[1em] \Rightarrow m' = \dfrac{5}{2}.$

Given, y-intercept = -3, putting values of slope and y-intercept in y = mx + c we get,

$\Rightarrow y = \dfrac{5}{2}x + (-3) \\[1em] \Rightarrow y = \dfrac{5x - 6}{2} \\[1em] \Rightarrow 2y = 5x - 6 \\[1em] \Rightarrow 5x - 2y - 6 = 0.$

Hence, the equation of the line is 5x - 2y - 6 = 0.

Find the equation of a straight line perpendicular to the line 3x - 4y + 12 = 0 and having same y-intercept as 2x - y + 5 = 0.

Answer

Given equation of line,

⇒ 3x - 4y + 12 = 0

Converting it in the form y = mx + c,

⇒ 4y = 3x + 12

⇒ y = $\dfrac{3}{4}x + 3$.

Comparing with y = mx + c we get,

Slope (m₁) = $\dfrac{3}{4}$.

Let the slope of the line perpendicular to the given line be m₂. So,

m₁ × m₂ = -1

$\Rightarrow \dfrac{3}{4} \times m_2 = -1 \\[1em] \Rightarrow m_2 = -\dfrac{4}{3}.$

The other line is 2x - y + 5 = 0 or y = 2x + 5.

Comparing with y= mx + c we get, c = 5.

So, the new line has slope = $-\dfrac{4}{3}$ and y-intercept = 5.

Putting these values in y = mx + c,

$\Rightarrow y = -\dfrac{4}{3}x + 5 \\[1em] \Rightarrow y = \dfrac{-4x + 15}{3} \\[1em] \Rightarrow 3y = -4x + 15 \\[1em] \Rightarrow 4x + 3y = 15 \\[1em] \Rightarrow 4x + 3y - 15 = 0.$

Hence, the equation of the line is 4x + 3y - 15 = 0.

Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15 = 0.

Answer

Given equation of line,

⇒ 3x + 5y + 15 = 0

Converting it in the form y = mx + c,

⇒ 5y = -3x - 15

⇒ y = $-\dfrac{3}{5}x - 3$

Comparing with y = mx + c we get,

Slope = $-\dfrac{3}{5}$

The slope of the line parallel to the given line will also be $-\dfrac{3}{5}$.

Given, the new line has slope = $-\dfrac{3}{5}$ and passes through (0, 4).

So, equation can be given by,

⇒ y - y₁ = m(x - x₁)

$\Rightarrow y - 4 = -\dfrac{3}{5}(x - 0) \\[1em] \Rightarrow 5(y - 4) = -3x \\[1em] \Rightarrow 5y - 20 = -3x \\[1em] \Rightarrow 3x + 5y - 20 = 0,$

Hence, the equation of the line is 3x + 5y - 20 = 0.

(i) The line 4x - 3y + 12 = 0 meets the x-axis at A. Write down the coordinates of A.

(ii) Determine the equation of the line passing through A and perpendicular to 4x - 3y + 12 = 0.

Answer

(i) When the line meets x-axis, its y-coordinate = 0.

So, putting y = 0 in 4x - 3y + 12 = 0, we get

⇒ 4x - 3(0) + 12 = 0
⇒ 4x = -12
⇒ x = -3.

Hence, the line meets the x-axis at A(-3, 0).

(ii) Converting 4x - 3y + 12 = 0, in the form y = mx + c.

⇒ 4x - 3y + 12 = 0

⇒ 3y = 4x + 12

⇒ y = $\dfrac{4}{3}x + 4$

Comparing the above equation with y = mx + c we get,

Slope (m₁) = $\dfrac{4}{3}$

Let the slope of the line perpendicular to the given line be m₂.

∴ m₁ × m₂ = -1

$\Rightarrow \dfrac{4}{3} \times m_2 = -1 \\[1em] \Rightarrow m_2 = -\dfrac{3}{4}.$

Equation of the line having slope = $-\dfrac{3}{4}$ and passing through (-3, 0) can be given by,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 0 = -\dfrac{3}{4}(x - (-3)) \\[1em] \Rightarrow 4y = -3(x + 3) \\[1em] \Rightarrow 4y = -3x - 9 \\[1em] \Rightarrow 4y + 3x + 9 = 0.$

Hence, the equation of the line is 3x + 4y + 9 = 0.

Find the equation of the line that is parallel to 2x + 5y - 7 = 0 and passes through the mid-point of the line segment joining the points (2, 7) and (-4, 1).

Answer

Given equation of line,

⇒ 2x + 5y - 7 = 0

Converting it in the form y = mx + c,

⇒ 5y = -2x + 7

⇒ y = $-\dfrac{2}{5}x + \dfrac{7}{5}$

So, the slope is $-\dfrac{2}{5}$.

Since, slope of parallel lines are equal. So, slope of parallel line will be $-\dfrac{2}{5}$

By mid-point formula, the mid-point of the line segment joining the points (2, 7) and (-4, 1) is

$\Big(\dfrac{2 + (-4)}{2}, \dfrac{7 + 1}{2}\Big)$ = (-1, 4).

Equation of the line having slope = $-\dfrac{2}{5}$ and passing through (-1, 4) can be given by,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 4 = -\dfrac{2}{5}(x - (-1)) \\[1em] \Rightarrow 5(y - 4) = -2(x + 1) \\[1em] \Rightarrow 5y - 20 = -2x - 2 \\[1em] \Rightarrow 5y + 2x = 20 - 2 \\[1em] \Rightarrow 2x + 5y = 18 \\[1em] \Rightarrow 2x + 5y - 18 = 0.$

Hence, the equation of the line is 2x + 5y - 18 = 0.

Find the equation of the line that is perpendicular to 3x + 2y - 8 = 0 and passes through the mid-point of the line segment joining the points (5, -2) and (2, 2).

Answer

Given equation of line,

⇒ 3x + 2y - 8 = 0

Converting it in the form y = mx + c,

⇒ 2y = -3x + 8

⇒ y = $-\dfrac{3}{2}x + 4$

Comparing with y = mx + c we get,

Slope (m₁) = $-\dfrac{3}{2}$

Now, the coordinates of the mid-point of the line segment joining the points (5, -2) and (2, 2) will be

$\Rightarrow \Big(\dfrac{5 + 2}{2}, \dfrac{-2 + 2}{2}\Big) \\[1em] \Rightarrow \Big(\dfrac{7}{2}, 0\Big).$

Let's consider the slope of the line perpendicular to the given line be m₂.

Then,

$\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow -\dfrac{3}{2} \times m_2 = -1 \\[1em] \Rightarrow m_2 = \dfrac{2}{3}.$

The equation of the new line with slope m₂ and passing through $(\dfrac{7}{2}, 0)$ can be given by point-slope form i.e.,

y - y₁ = m(x - x₁)

Putting values we get,

$\Rightarrow y - 0 = \dfrac{2}{3}(x - \dfrac{7}{2}) \\[1em] \Rightarrow 3y = 2(x - \dfrac{7}{2}) \\[1em] \Rightarrow 3y = 2x - 7 \\[1em] \Rightarrow 2x - 3y - 7 = 0.$

Hence, the equation of the line is 2x - 3y - 7 = 0.

Find the equation of a straight line passing through the intersection of 2x + 5y - 4 = 0 with x-axis and parallel to the line 3x - 7y + 8 = 0.

Answer

Let the point of intersection of the line 2x + 5y - 4 = 0 and the x-axis be (x₁, 0).

Substituting the value of points in equation,

⇒ 2x₁ + 5 × 0 - 4 = 0
⇒ 2x₁ = 4
⇒ x₁ = 2.

Coordinates of the point of intersection will be (2, 0).

Given new line is parallel to 3x - 7y + 8 = 0.

Converting it in the form y = mx + c,

3x - 7y + 8 = 0

⇒ 7y = 3x + 8

⇒ y = $\dfrac{3}{7}x + \dfrac{8}{7}$

Comparing equation with y = mx + c, we get slope = $\dfrac{3}{7}$.

The equation of the line with slope $\dfrac{3}{7}$ and passing through (2, 0) can be given by point-slope form,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 0 = \dfrac{3}{7}(x - 2) \\[1em] \Rightarrow 7y = 3x - 6 \\[1em] \Rightarrow 3x - 7y - 6 = 0.$

Hence, the equation of the new line is 3x - 7y - 6 = 0.

Line AB is perpendicular to line CD. Coordinates of B, C and D are (4, 0), (0, -1) and (4, 3) respectively. Find

(i) the slope of CD

(ii) the equation of line AB

Answer

(i) By formula,

Slope of a line = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

Slope of CD = $\dfrac{3 - (-1)}{4 - 0} = \dfrac{4}{4}$ = 1.

Hence, slope of CD = 1.

(ii) We know that,

The product of slope of two perpendicular lines equals to -1.

∴ Slope of AB × Slope of CD = -1

⇒ Slope of AB × 1 = -1

⇒ Slope of AB = -1.

By point-slope formula,

Equation of line :

⇒ y - y₁ = m(x - x₁)

Equation of AB :

⇒ y - 0 = -1(x - 4)

⇒ y = -x + 4

⇒ x + y = 4.

Hence, equation of AB is x + y = 4.

Find the equation of a line parallel to the line 2x + y - 7 = 0 and passing through the point of intersection of the lines x + y - 4 = 0 and 2x - y = 8.

Answer

Simultaneously solving equations :

⇒ x + y - 4 = 0 .......(1)

⇒ 2x - y = 8 ........(2)

Solving equation (1), we get :

⇒ x = 4 - y ...........(3)

Substituting value of x from (3) in (2), we get :

⇒ 2(4 - y) - y = 8

⇒ 8 - 2y - y = 8

⇒ 8 - 3y = 8

⇒ 3y = 0

⇒ y = 0.

Substituting value of y in (3), we get :

⇒ x = 4 - 0 = 4.

Point of intersection = (4, 0).

Given,

Equation :

⇒ 2x + y - 7 = 0

⇒ y = -2x + 7

Comparing above equation with y = mx + c, we get :

⇒ m = -2.

We know that,

Slope of parallel lines are equal.

∴ Slope of line parallel to 2x + y - 7 is -2.

By point-slope formula,

Equation of line :

⇒ y - y₁ = m(x - x₁)

Substituting value we get :

Equation of line parallel to line 2x + y - 7 = 0 and passing through the point of intersection of the lines x + y - 4 = 0 and 2x - y = 8 is :

⇒ y - 0 = -2(x - 4)

⇒ y = -2x + 8

⇒ 2x + y = 8.

Hence, the equation of required line is 2x + y = 8.

The equation of a line is 3x + 4y - 7 = 0. Find

(i) slope of the line.

(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x - y + 2 = 0 and 3x + y - 10 = 0.

Answer

(i) Given, 3x + 4y - 7 = 0

Converting the equation in the form of y = mx + c,

⇒ 4y = -3x + 7

⇒ y = $-\dfrac{3}{4}x + \dfrac{7}{4}$

Comparing the equation with y = mx + c, we get slope (m₁) = $-\dfrac{3}{4}$.

(ii) Let the slope of the line perpendicular to the given line be m₂.

Then,

$\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow -\dfrac{3}{4} \times m_2 = -1 \\[1em] \Rightarrow m_2 = \dfrac{4}{3}.$

Now to find the point of intersection of

x - y + 2 = 0 ... (i)
3x + y - 10 = 0 ... (ii)

On adding (i) and (ii), we get

⇒ x - y + 2 + 3x + y - 10 = 0
⇒ 4x - 8 = 0
⇒ 4x = 8
⇒ x = 2.

Putting x = 2 in (i), we get

⇒ 2 - y + 2 = 0
⇒ y = 4.

Hence, the point of intersection of the lines is (2, 4).

The equation of the line with slope $\dfrac{4}{3}$ and passing through (2, 4) can be given by point-slope form,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 4 = \dfrac{4}{3}(x - 2) \\[1em] \Rightarrow 3(y - 4) = 4(x - 2) \\[1em] \Rightarrow 3y - 12 = 4x - 8 \\[1em] \Rightarrow 4x - 3y + 4 = 0.$

Hence, the equation of the new line is 4x - 3y + 4 = 0.

Find the equation of the perpendicular from the point (1, -2) on the line 4x - 3y - 5 = 0. Also find the coordinates of the foot of perpendicular.

Answer

Converting 4x - 3y - 5 = 0 in the form of y = mx + c.

⇒ 4x - 3y - 5 = 0

⇒ 3y = 4x - 5

⇒ y = $\dfrac{4}{3}x - \dfrac{5}{3}$

Slope of the line (m₁) = $\dfrac{4}{3}$.

Let the slope of the line perpendicular to 4x - 3y - 5 = 0 be m₂.

Then, m₁ × m₂ = -1.

$\Rightarrow \dfrac{4}{3} \times m_2 = -1 \\[1em] \Rightarrow m_2 = -\dfrac{3}{4}.$

The equation of the line having slope m₂ and passing through the point (1, -2) can be given by point-slope form i.e.,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - (-2) = -\dfrac{3}{4}(x - 1) \\[1em] \Rightarrow 4(y + 2) = -3(x - 1) \\[1em] \Rightarrow 4y + 8 = -3x + 3 \\[1em] \Rightarrow 3x + 4y + 5 = 0$

For finding the coordinates of the foot of the perpendicular which is the point of intersection of the lines

4x - 3y - 5 = 0 ....(i)
3x + 4y + 5 = 0 ....(ii)

On multiplying (i) by 4 and (ii) by 3 we get,

16x - 12y - 20 = 0 ....(iii)
9x + 12y + 15 = 0 ....(iv)

Adding (iii) and (iv) we get,

⇒ 16x - 12y - 20 + 9x + 12y + 15 = 0

⇒ 25x - 5 = 0

⇒ x = $\dfrac{5}{25}$

⇒ x = $\dfrac{1}{5}$.

Putting value of x in (i), we have

$\Rightarrow 4 \times \dfrac{1}{5} - 3y - 5 = 0 \\[1em] \Rightarrow \dfrac{4}{5} - 3y - 5 = 0 \\[1em] \Rightarrow 3y = \dfrac{4}{5} - 5 \\[1em] \Rightarrow 3y = \dfrac{4 - 25}{5} \\[1em] \Rightarrow 3y = -\dfrac{21}{5} \\[1em] \Rightarrow y = -\dfrac{7}{5}.$

∴ Coordinates = $\Big(\dfrac{1}{5}, -\dfrac{7}{5}\Big)$.

Hence, the equation of the new line is 3x + 4y + 5 = 0 and coordinates of the foot of perpendicular (i.e., its intersection with 4x - 3y - 5 = 0) are $\Big(\dfrac{1}{5}, -\dfrac{7}{5}\Big)$.

Prove that the line through (0, 0) and (2, 3) is parallel to the line through (2, -2) and (6, 4).

Answer

The slope of the line passing through two points (x₁, y₁) and (x₂, y₂) is given by

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$.

So, slope (m₁) of (0, 0) and (2, 3) is,

$= \dfrac{3 - 0}{2 - 0} \\[1em] = \dfrac{3}{2}.$

So, slope (m₂) of (2, -2) and (6, 4) is,

$= \dfrac{4 - (-2)}{6 - 2} \\[1em] = \dfrac{6}{4} \\[1em] = \dfrac{3}{2}.$

Since, m₁ = $\dfrac{3}{2}$ = m₂.

Hence, the lines are parallel to each other.

Prove that the line through, (-2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (4, 24).

Answer

The slope of the line passing through two points (x₁, y₁) and (x₂, y₂) is given by

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$.

Slope (m₁) of line joining (-2, 6) and (4, 8) is,

$= \dfrac{8 - 6}{4 - (-2)} \\[1em] = \dfrac{2}{6} \\[1em] = \dfrac{1}{3}.$

Slope (m₂) of line joining (8, 12) and (4, 24) is,

$= \dfrac{24 - 12}{4 - 8} \\[1em] = \dfrac{12}{-4} \\[1em] = -3.$

Since, m₁ × m₂ = $\dfrac{1}{3} \times -3 = -1$.

Hence, the lines are perpendicular to each other.

Show that the triangle formed by the points A(1, 3), B(3, -1) and C(-5, -5) is a right angled triangle (by using slopes).

Answer

The slope of the line passing through two points (x₁, y₁) and (x₂, y₂) is given by

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$.

Slope (m₁) of the line joining A(1, 3) and B(3, -1) is,

$= \dfrac{-1 - 3}{3 - 1} \\[1em] = \dfrac{-4}{2} \\[1em] = -2.$

Slope (m₂) of line joining B(3, -1) and C(-5, -5) is,

$= \dfrac{-5 - (-1)}{-5 - 3} \\[1em] = \dfrac{-4}{-8} \\[1em] = \dfrac{1}{2}.$

Since, m₁ × m₂ = $-2 \times \dfrac{1}{2} = -1$.

Thus AB and BC are perpendicular to each other.

Hence, △ABC is a right-angled triangle.

Find the equation of the line through the point (-1, 3) and parallel to the line joining the points (0, -2) and (4, 5).

Answer

The slope of the line passing through two points (x₁, y₁) and (x₂, y₂) is given by

Slope = $\dfrac{y_2 - y_1}{x_2 - x_1}$.

Slope of the line joining the points (0, -2) and (4, 5) is,

$= \dfrac{5 + 2}{4 - 0} \\[1em] = \dfrac{7}{4}.$

∴ Slope of line parallel to the line joining (0, -2) and (4, 5) = $\dfrac{7}{4}$

The equation of the line having slope $\dfrac{7}{4}$ and passing through (-1, 3) can be given by point-slope form i.e.,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 3 = \dfrac{7}{4}(x - (-1)) \\[1em] \Rightarrow 4(y - 3) = 7(x + 1) \\[1em] \Rightarrow 4y - 12 = 7x + 7 \\[1em] \Rightarrow 7x - 4y + 19 = 0.$

Hence, the equation of the line through the point (-1, 3) and parallel to the line joining the points (0, -2) and (4, 5) is 7x - 4y + 19 = 0.

A(-1, 3), B(4, 2), C(3, -2) are the vertices of a triangle.

(i) Find the coordinates of the centroid G of the triangle.

(ii) Find the equation of the line through G and parallel to AC.

Answer

(i) Centroid of the triangle is given by,

$G(x, y) = \Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big) \\[1em] = \Big(\dfrac{-1 + 4 + 3}{3}, \dfrac{3 + 2 - 2}{3}\Big) \\[1em] = \Big(\dfrac{6}{3}, \dfrac{3}{3}\Big) \\[1em] = (2, 1).$

Hence, the coordinates of the centroid G of the triangle is (2, 1).

(ii) Slope of AC = $\dfrac{y_2 - y_1}{x_2 - x_1}$

$= \dfrac{-2 - 3}{3 - (-1)} \\[1em] = -\dfrac{5}{4}.$

So, the slope of the line parallel to AC is also $-\dfrac{5}{4}.$ and it passes through (2, 1). Hence, its equation can be given by point-slope form i.e.,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 1 = -\dfrac{5}{4}(x - 2) \\[1em] \Rightarrow 4(y - 1) = -5(x - 2) \\[1em] \Rightarrow 4y - 4 = -5x + 10 \\[1em] \Rightarrow 4y + 5x = 14 \\[1em] \Rightarrow 5x + 4y - 14 = 0.$

Hence, the equation of the required line is 5x + 4y - 14 = 0.

Find the equation of the line through (0, -3) and perpendicular to the line joining the points (-3, 2) and (9, 1).

Answer

The slope (m₁) of the line joining (-3, 2) and (9, 1) i.e. two points is $\dfrac{y_2 - y_1}{x_2 - x_1}$ so,

$\text{m}_1 = \dfrac{1 - 2}{9 - (-3)} \\[1em] = -\dfrac{1}{12}.$

Let the slope of the line perpendicular to the above line be m₂.

Then, m₁ × m₂ = -1.

$\Rightarrow -\dfrac{1}{12} \times m_2 = -1 \\[1em] \Rightarrow m_2 = 12.$

So, the equation of the line passing through (0, -3) and slope 12 can be given by point-slope form i.e.,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - (-3) = 12(x - 0) \\[1em] \Rightarrow y + 3 = 12x \\[1em] \Rightarrow 12x - y - 3 = 0.$

Hence, the equation of the required line is 12x - y - 3 = 0.

The vertices of a △ABC are A(3, 8), B(-1, 2) and C(6, -6). Find:

(i) slope of BC.

(ii) equation of a line perpendicular to BC and passing through A.

Answer

(i) Let the slope of BC be m₁. Slope of BC is given by,

$\text{m}_1 = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{-6 - 2}{6 - (-1)} \\[1em] = -\dfrac{8}{7}.$

Hence, the slope of BC is $-\dfrac{8}{7}.$

(ii) Let slope of line perpendicular to BC be m₂.

So, m₁ × m₂ = -1.

$\Rightarrow -\dfrac{8}{7} \times m_2 = -1 \\[1em] \Rightarrow m_2 = \dfrac{7}{8}.$

Equation of the line having the slope = $\dfrac{7}{8}$ and passing through A(3, 8) can be given by point-slope formula i.e.,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 8 = \dfrac{7}{8}(x - 3) \\[1em] \Rightarrow 8(y - 8) = 7(x - 3) \\[1em] \Rightarrow 8y - 64 = 7x - 21 \\[1em] \Rightarrow 7x - 8y - 21 + 64 = 0 \\[1em] \Rightarrow 7x - 8y + 43 = 0.$

Hence, the equation of the required line is 7x - 8y + 43 = 0.

The vertices of a triangle are A(10, 4), B(4, -9) and C(-2, -1). Find the equation of the altitude through A.
[The perpendicular drawn from a vertex of a triangle to the opposite side is called altitude.]

Answer

Given, vertices of a triangle are A(10, 4), B(4, -9) and C(-2, -1).

Now,

Slope of line BC (m₁),

$\text{m}_1 = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{-1 - (-9)}{-2 - 4} \\[1em] = -\dfrac{8}{6} \\[1em] = -\dfrac{4}{3}.$

Let the slope of the altitude from A(10, 4) to BC be m₂.

Then, m₁ × m₂ = -1.

$\Rightarrow -\dfrac{4}{3} \times m_2 = -1 \\[1em] \Rightarrow m_2 = \dfrac{3}{4}.$

Equation of the line having the slope = $\dfrac{3}{4}$ and passing through A(10, 4) can be given by point-slope formula i.e.,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 4 = \dfrac{3}{4}(x - 10) \\[1em] \Rightarrow 4(y - 4) = 3(x - 10) \\[1em] \Rightarrow 4y - 16 = 3x - 30 \\[1em] \Rightarrow 3x - 4y + 16 - 30 = 0 \\[1em] \Rightarrow 3x - 4y - 14 = 0.$

Hence, the equation of the required line is 3x - 4y - 14 = 0.

A(2, -4), B(3, 3) and C(-1, 5) are the vertices of triangle ABC. Find the equation of :

(i) the median of the triangle through A.

(ii) the altitude of the triangle through B.

Answer

Triangle ABC with vertices A(2, -4), B(3, 3) and C(-1, 5) is shown below:

(i) Let D be the mid-point of BC. So, AD will be the median.

Coordinates of D by mid-point formula will be,

$= \Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) \\[1em] = \Big(\dfrac{3 + (-1)}{2}, \dfrac{3 + 5}{2}\Big) \\[1em] = \Big(\dfrac{2}{2}, \dfrac{8}{2}\Big) \\[1em] = (1, 4).$

The equation of AD can be given by two-point formula i.e.,

$\Rightarrow y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1) \\[1em] \Rightarrow y - (-4) = \dfrac{4 - (-4)}{1 -2}(x - 2) \\[1em] \Rightarrow y + 4 = -8(x - 2) \\[1em] \Rightarrow y + 4 = -8x + 16 \\[1em] \Rightarrow 8x + y - 12 = 0.$

Hence, the equation of the median of the triangle through A is 8x + y - 12 = 0.

(ii) Let E be a point on AC such that BE is perpendicular to AC.

Slope (m₁) of AC is,

$\text{m}_1 = \dfrac{5 - (-4)}{-1 - 2} \\[1em] = -\dfrac{9}{3} \\[1em] = -3.$

Let slope of BE be m₂. Since, BE is perpendicular to AC so,

$\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow -3 \times m_2 = -1 \\[1em] \Rightarrow m_2 = \dfrac{1}{3}.$

So, the equation of BE by point-slope form will be

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 3 = \dfrac{1}{3}(x - 3) \\[1em] \Rightarrow 3(y - 3) = x - 3 \\[1em] \Rightarrow 3y - 9 = x - 3 \\[1em] \Rightarrow x - 3y + 6 = 0.$

Hence, the equation of the required line is x - 3y + 6 = 0.

Find the equation of the right bisector of the line segment joining the points (1, 2) and (5, -6).

Answer

Slope of the line joining the points (1, 2) and (5, -6) is,

$\text{m}_1 = \dfrac{-6 - 2}{5 - 1} \\[1em] = -\dfrac{8}{4} \\[1em] = -2.$

Let m₂ be the slope of the right bisector of the above line. Then,

$\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow -2 \times m_2 = -1 \\[1em] \Rightarrow m_2 = \dfrac{1}{2}.$

The mid-point of the line segment joining (1, 2) and (5, -6) will be

$=\Big(\dfrac{1 + 5}{2}, \dfrac{2 + (-6)}{2}\Big) \\[1em] = (3, -2).$

Equation of the line having the slope = $\dfrac{1}{2}$ and passing through (3, -2) can be given by point-slope formula i.e.,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - (-2) = \dfrac{1}{2}(x - 3) \\[1em] \Rightarrow 2(y + 2) = x - 3 \\[1em] \Rightarrow 2y + 4 = x - 3 \\[1em] \Rightarrow x - 2y - 3 - 4 = 0 \\[1em] \Rightarrow x - 2y - 7 = 0.$

Hence, the equation of the required right bisector is x - 2y - 7 = 0.

Points A and B have coordinates (7, -3) and (1, 9) respectively. Find

(i) the slope of AB.

(ii) the equation of the perpendicular bisector of the line segment AB.

(iii) the value of p if (-2, p) lies on it.

Answer

(i) Slope (m₁) of AB is,

$\text{m}_1 = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{9 - (-3)}{1 - 7} \\[1em] = -\dfrac{12}{6} \\[1em] = -2.$

Hence, the slope of AB is -2.

(ii) Let PQ be the perpendicular bisector of AB intersecting it at M. Now, the coordinates of M will be

$= \Big(\dfrac{7 + 1}{2}, \dfrac{-3 + 9}{2}\Big) \\[1em] = (4, 3).$

Let the slope of the line PQ be m₂. Since, PQ is perpendicular to AB then product of their slopes will be equal to -1,

$\therefore m_1 \times m_2 = -1 \\[1em] \Rightarrow -2 \times m_2 = -1 \\[1em] \Rightarrow m_2 = \dfrac{1}{2}.$

Thus, by point-slope form equation of PQ is,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 3 = \dfrac{1}{2}(x - 4) \\[1em] \Rightarrow 2(y - 3) = x - 4 \\[1em] \Rightarrow 2y - 6 = x - 4 \\[1em] \Rightarrow x - 2y + 2 = 0.$

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

(iii) As (-2, p) lies on the above line. The point will satisfy the line equation x - 2y + 2 = 0.

⇒ -2 - 2p + 2 = 0
⇒ 2p = 0
⇒ p = 0.

Hence, the value of p is 0.

The points B(1, 3) and D(6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.

Answer

Slope of BD is given by

m₁ = $\dfrac{8 - 3}{6 - 1} = \dfrac{5}{5}$ = 1.

We know that diagonal AC is a perpendicular bisector of diagonal BD.

So, the slope of AC (m₂) will be,

$\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow 1 \times m_2 = -1 \\[1em] \Rightarrow m_2 = -1.$

Coordinates of mid-point of BD and AC will be same as diagonals of a square meet at their mid-point.

$= \Big(\dfrac{1 + 6}{2}, \dfrac{3 + 8}{2}\Big) \\[1em] = \Big(\dfrac{7}{2}, \dfrac{11}{2}\Big).$

By point-slope formula equation of AC is

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - \dfrac{11}{2} = -1(x - \dfrac{7}{2}) \\[1em] \Rightarrow y - \dfrac{11}{2} = -x + \dfrac{7}{2} \\[1em] \Rightarrow y + x -\dfrac{11}{2} - \dfrac{7}{2} = 0 \\[1em] \Rightarrow y + x - \dfrac{18}{2} = 0 \\[1em] \Rightarrow x + y - 9 = 0.$

Hence, the equation of the required line is x + y - 9 = 0.

ABCD is a rhombus. The coordinates of A and C are (3, 6) and (-1, 2) respectively. Write down the equation of BD.

Answer

Slope of AC is given by,

$m_1 = \dfrac{2 - 6}{-1 - 3} = \dfrac{-4}{-4}$ = 1.

We know that diagonal of a rhombus bisect each other at right angles. So, the diagonal BD is perpendicular to diagonal AC.

Let the slope of BD be m₂. Then,

$\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow 1 \times m_2 = -1 \\[1em] \Rightarrow m_ 2 = -1.$

Coordinates of mid-point of AC and BD are same which are

$= \Big(\dfrac{3 + (-1)}{2}, \dfrac{6 + 2}{2}\Big) \\[1em] = \Big(\dfrac{2}{2}, \dfrac{8}{2}\Big) \\[1em] = (1, 4).$

Equation of the line having the slope = -1 and passing through (1, 4) can be given by point-slope formula i.e.,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 4 = -1(x - 1) \\[1em] \Rightarrow y - 4 = -x + 1 \\[1em] \Rightarrow x + y - 4 - 1 = 0 \\[1em] \Rightarrow x + y - 5 = 0.$

Hence, the equation of BD is x + y - 5 = 0.

Find the image of the point (1, 2) in the line x - 2y - 7 = 0.

Answer

The given line is x - 2y - 7 = 0 .....(i)

⇒ 2y = x - 7

⇒ y = $\dfrac{1}{2}x - \dfrac{7}{2}$.

The slope of the line (i) = m₁ = $\dfrac{1}{2}.$

Let the point (1, 2) be P.

From P draw a perpendicular to the line (i) and produce it to point P' such that P'M = MP, then P' is the image of P in line (i) and line (i) is the right bisector of the segment PP'.

Let P' be (a, b).

Then slope of PP' = m₂ = $\dfrac{b - 2}{a - 1}$.

Since, line (i) is perpendicular to PP' so,

$\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{b - 2}{a - 1} = -1 \\[1em] \Rightarrow \dfrac{b - 2}{2a - 2} = -1 \\[1em] \Rightarrow b - 2 = -2a + 2 \\[1em] \Rightarrow 2a + b = 4 \space .....(\text{iii})$

Also mid-point of PP' is M$\Big(\dfrac{a + 1}{2}, \dfrac{b + 2}{2}\Big)$.

Since, (i) is the right bisector of the segment PP', M lies on (i)

$\Rightarrow \dfrac{a + 1}{2} - 2\Big(\dfrac{b + 2}{2}\Big) - 7 = 0 \\[1em] \Rightarrow \dfrac{a + 1}{2} - b - 2 - 7 = 0 \\[1em] \Rightarrow \dfrac{a + 1}{2} - b = 9 \\[1em] \Rightarrow \dfrac{a + 1 - 2b}{2} = 9 \\[1em] \Rightarrow a + 1 - 2b = 18 \\[1em] \Rightarrow a - 2b = 17 \space ....(\text{iv})$

Multiplying equation (iv) by 2 and subtracting from (iii) we get,

$\Rightarrow 2a + b - 2(a - 2b) = 4 - 34 \\[1em] \Rightarrow 2a + b - 2a + 4b = -30 \\[1em] \Rightarrow 5b = -30 \\[1em] \Rightarrow b = -6.$

Putting value of b in Eq (iii),

⇒ 2a - 6 = 4
⇒ 2a = 10
⇒ a = 5.

P' = (a, b) = (5, -6).

Hence, the coordinates of image are (5, -6).

If the line x - 4y - 6 = 0 is the perpendicular bisector of the line segment PQ and the coordinates of P are (1, 3), find the coordinates of Q.

Answer

Given, equation of line,

⇒ x - 4y - 6 = 0

⇒ 4y = x - 6

⇒ y = $\dfrac{1}{4}x - \dfrac{6}{4}.$

Comparing with y = mx + c we get, slope = $\dfrac{1}{4}$.

Since, given line and PQ are perpendicular so their products will be equal to -1. Let slope of PQ be m₁,

$\therefore \dfrac{1}{4} \times m_1 = -1 \\[1em] \Rightarrow m_1 = -4.$

Hence, slope of PQ = -4.

Now equation of PQ can be found by point slope form i.e.,

$\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 3 = -4(x - 1) \\[1em] \Rightarrow y - 3 = -4x + 4 \\[1em] \Rightarrow 4x + y - 7 = 0,$

Since, line x - 4y - 6 = 0 is perpendicular bisector of 4x + y - 7 = 0 hence solving them simultaneously to find point of intersection,

⇒ x - 4y = 6 ......(i)
⇒ 4x + y = 7 ......(ii)

Multiplying (ii) with 4 and adding with (i) we get,

⇒ 16x + 4y + x - 4y = 28 + 6
⇒ 17x = 34
⇒ x = 2.

Putting value of x = 2 in (i),

⇒ 2 - 4y = 6
⇒ -4y = 4
⇒ y = -1.

Hence, the point of intersection which is the mid-point of PQ is (2, -1).

Let coordinates of Q be (a, b).

By mid-point formula, coordinates of mid-point of PQ are

$\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big) \\[1em] = \Big(\dfrac{1 + a}{2}, \dfrac{3 + b}{2}\Big) \\[1em]$

Equating with mid-point of PQ (2, -1) we get,

$2 = \dfrac{1 + a}{2} \text{ and } -1 = \dfrac{3 + b}{2} \\[1em] 4 = 1 + a \text{ and } -2 = 3 + b \\[1em] a = 4 - 1 \text{ and } b = -2 - 3 \\[1em] a = 3 \text{ and } b = -5$

Hence, the coordinates of Q are (3, -5).

OABC is a square, O is the origin and the points A and B are (3, 0) and (p, q). If OABC lies in the first quadrant, find the values of p and q. Also write down the equations of AB and BC.

Answer

The square OABC is plotted on the graph below:

$\text{OA} = \sqrt{(3 - 0)^2 + (0 - 0)^2} \\[1em] = \sqrt{3^2 + 0^2} \\[1em] = \sqrt{9} \\[1em] = 3. \\[1em] \text{AB} = \sqrt{(3 - p)^2 + (0 - q)^2} \\[1em] = \sqrt{(3 - p)^2 + q^2}$

Since, OA = AB (as sides of square are equal)

$\therefore \sqrt{(p - 3)^2 + q^2} = 3 \\[1em] \Rightarrow (p - 3)^2 + q^2 = 9 \\[1em] \Rightarrow p^2 + 9 - 6p + q^2 = 9 \\[1em] \Rightarrow p^2 + q^2 - 6p = 0 \space .....(\text{i})$

By pythagoras theorem, OB² = OA² + AB².

$\Rightarrow \Big(\sqrt{(p - 0)^2 + (q - 0)^2}\Big)^2 = 3^2 + \Big(\sqrt{(3 - p)^2 + q^2}\Big)^2 \\[1em] \Rightarrow p^2 + q^2 = 9 + (3 - p)^2 + q^2 \\[1em] \Rightarrow p^2 + q^2 = 9 + 9 + p^2 - 6p + q^2 \\[1em] \Rightarrow p^2 - p^2 + q^2 - q^2 + 6p = 18 \\[1em] \Rightarrow 6p = 18 \\[1em] \Rightarrow p = 3.$

Substituting value of p in (i),

$\Rightarrow 3^2 + q^2 - 6(3) = 0 \\[1em] \Rightarrow q^2 + 9 - 18 = 0 \\[1em] \Rightarrow q^2 - 9 = 0 \\[1em] \Rightarrow q^2 - (3)^2 = 0 \\[1em] \Rightarrow (q - 3)(q + 3) = 0 \\[1em] \Rightarrow q - 3 = 0 \text{ or } q + 3 = 0 \\[1em] \Rightarrow q = 3 \text{ or } q = -3 \\[1em] \Rightarrow q = 3, -3.$

But q = -3 is not possible as the square is in 1st quadrant and the coordinates are positive in 1st quadrant.

∴ p = 3 and q = 3.

AB is parallel to y-axis,

∴ Equation of AB will be x = 3 or x - 3 = 0.

BC is parallel to x-axis,

∴ Equation BC will be y = 3 or y - 3 = 0.

Hence, the value of p = 3 and q = 3. Equation of AB is x - 3 = 0 and BC is y - 3 = 0.