Equation of A Straight Line

Find the slope of the line passing through the points:

(i) A(–2, 1) and B(3, –4)

(ii) A(0, –3) and B(2, 1)

(iii) A(4, –9) and B(–2, –1)

(iv) A(2, 5) and B(–4, –4)

Answer

(i) A(-2, 1) and B(3, -4)

By formula,

Slope (m) = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

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

Hence, slope is -1.

(ii) A(0, –3) and B(2, 1)

By formula,

Slope (m) = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

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

Hence, slope is 2.

(iii) A(4, –9) and B(–2, –1)

By formula,

Slope (m) = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\Rightarrow m = \dfrac{-1 - (-9)}{-2 - 4} \\[1em] = \dfrac{-1 + 9}{-6} \\[1em] = \dfrac{8}{-6} \\[1em] = -\dfrac{4}{3}.$

Hence, slope is $-\dfrac{4}{3}$.

(iv) A(2, 5) and B(–4, –4)

By formula,

Slope (m) = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\Rightarrow m = \dfrac{-4 - 5}{-4 - 2} \\[1em] = \dfrac{-9}{-6} \\[1em] = \dfrac{3}{2}.$

Hence, slope is $\dfrac{3}{2}$.

If the slope of the line joining P(k, 3) and Q(8, –6) is $\dfrac{-3}{4}$, find the value of k.

Answer

Given,

P(k, 3) and Q(8, –6)

Slope = $\dfrac{-3}{4}$

By formula,

Slope (m) = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\Rightarrow \text{Slope of PQ} = \dfrac{-6 - 3}{8 - k} \\[1em] \Rightarrow -\dfrac{3}{4} = \dfrac{-6 - 3}{8 - k} \\[1em] \Rightarrow -\dfrac{3}{4} = \dfrac{-9}{8 - k} \\[1em] \Rightarrow 3(8 - k) = 4(9) \\[1em] \Rightarrow 24 -3k = 36 \\[1em] \Rightarrow -3k = 36 - 24 \\[1em] \Rightarrow -3k = 12 \\[1em] \Rightarrow k = \dfrac{12}{-3} \\[1em] \Rightarrow k = -4.$

Hence, k = -4.

Without using the distance formula, prove that the points A(1, 4), B(3, –2) and C(–3, 16) are collinear.

Answer

To prove that the points A(1, 4), B(3, –2), and C(–3, 16) are collinear we must show that the slope between any pair of points is the same.

By formula,

Slope (m) = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\Rightarrow \text{Slope of AB} = \dfrac{-2 - 4}{3 - 1} \\[1em] = \dfrac{-6}{2} \\[1em] = -3. \\[1em] \Rightarrow \text{Slope of BC} = \dfrac{16 -(-2)}{-3 - 3} \\[1em] = \dfrac{16 + 2}{-6} \\[1em] = \dfrac{18}{-6} \\[1em] = -3.$

Slope of AB = Slope of BC

Hence, proved points A, B and C are collinear.

Find the value of k such that the points P(k, 1), Q(2, –5) and R(k - 2, –3) are collinear.

Answer

Given,

Points P, Q and R are collinear.

Thus, the slope of PQ equal to the slope of QR.

By formula,

Slope (m) = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$\Rightarrow \text{Slope of PQ} = \dfrac{-5 - 1}{2 - k} \\[1em] = \dfrac{-6}{2 - k} \\[1em] \Rightarrow \text{Slope of QR} = \dfrac{-3 -(-5)}{(k - 2) - 2} \\[1em] = \dfrac{-3 + 5}{k - 4} \\[1em] = \dfrac{2}{k - 4}.$

Slope of PQ = Slope of QR

$\Rightarrow \dfrac{-6}{2 - k} = \dfrac{2}{k - 4}$

⇒ -6(k - 4) = 2(2 - k)

⇒ -6k + 24 = 4 - 2k

⇒ 24 - 4 = -2k + 6k

⇒ 20 = 4k

⇒ k = $\dfrac{20}{4}$

⇒ k = 5

Hence, k = 5.

Find the equation of a line parallel to the x-axis and passing through the point (–3, 2).

Answer

We know that the equation of straight line parallel to x-axis is

y = a

Since the line passes through the point (-3,2), we get

a = 2

∴ Equation of the line

⇒ y = 2 or y - 2 = 0.

Hence, equation of the line is y = 2.

Find the equation of a line parallel to the y-axis and passing through the point (–7, 5).

Answer

We know that the equation of straight line parallel to y-axis is x = a.

Since the line passes through the point (-7, 5), we get

a = -7

∴ Equation of the line

⇒ x = -7 or x + 7 = 0.

Hence, equation of the line is x + 7 = 0.

Find the equation of a line whose inclination is 30° and whose y-intercept is –2.

Answer

Given,

θ = 30° and c = -2.

We know that,

m = tan θ = tan 30° = $\dfrac{1}{\sqrt3}$.

Substituting values of m and c in equation y = mx + c, we get :

$\Rightarrow y = \dfrac{1}{\sqrt3} x + (-2) \\[1em] \Rightarrow \sqrt3y = x - 2\sqrt{3} \\[1em] \Rightarrow \sqrt3y - x + 2\sqrt{3} = 0.$

Hence, equation of the line is $\sqrt3y - x + 2\sqrt{3}$.

Find the equation of a line whose:

(i) Slope = $\dfrac{3}{4}$ and y-intercept = –4

(ii) Gradient = $\sqrt{3}$ and y-intercept = $\dfrac{-2}{3}$

Answer

(i) The equation of the straight line is given by, y = mx + c, we get where m is the slope and c is the y-intercept.

Given slope = $\dfrac{3}{4}$ and y-intercept = -4. Substituting values in equation we get,

⇒ y = $\dfrac{3}{4}$x - 4.

⇒ y = $\dfrac{3x - 16}{4}$

⇒ 4y = 3x - 16

⇒ 3x - 4y = 16.

Hence, equation of the line is 3x - 4y = 16.

(ii) The equation of straight line is given by y = mx + c, where m is the slope and c is the y-intercept.

Given slope = $\sqrt{3}$ and y-intercept = $\dfrac{-2}{3}$.

Substituting values in equation we get,

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

Hence, equation of the line is $3\sqrt{3}x -3y - 2 = 0$.

Find the equation of a line which makes an angle of 60° with the positive direction of the x-axis and passes through the point P(0, –3).

Answer

m = tan θ = tan 60° = $\sqrt3$

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get,

$\Rightarrow y - (-3) = \sqrt3(x - 0) \\[1em] \Rightarrow y + 3 = \sqrt3x \\[1em] \Rightarrow \sqrt3x - y = 3.$

Hence, equation of the line is $\sqrt3x - y = 3$.

Find the equation of a line:

(i) Whose slope is 4 and which passes through the point (3, 7)

(ii) Whose slope is –3 and which passes through the point (–2, 3)

Answer

(i) By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get,

⇒ y - 7 = 4(x - 3)

⇒ y - 7 = 4x - 12

⇒ y = 4x - 12 + 7

⇒ y = 4x - 5

⇒ 4x - y = 5.

Hence, equation of the line is 4x - y = 5.

(ii) By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get,

⇒ y - 3 = -3(x - (-2))

⇒ y - 3 = -3(x + 2)

⇒ y - 3 = -3x - 6

⇒ y = -3x - 6 + 3

⇒ y = -3x - 3

⇒ 3x + y + 3 = 0

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

Find the gradient and the y-intercept of each of the following lines:

(i) 5x – 10y = 3

(ii) $\dfrac{x}{6} + \dfrac{y}{9} = 1$

(iii) x + 4 = 0

(iv) y = 6

Answer

(i) Given,

⇒ 5x – 10y = 3

Converting 5x – 10y = 3 in the form y = mx + c we get,

⇒ -10y = 3 - 5x

⇒ y = $-\dfrac{3}{10} - \dfrac{5x}{-10}$

⇒ y = $-\dfrac{3}{10} + \dfrac{x}{2}$

⇒ y = $\dfrac{1}{2}x - \dfrac{3}{10}$

The equation of straight line is given by, y = mx + c, where m is the slope and c is the y-intercept.

Comparing, y = mx + c with y = $\dfrac{1}{2}x - \dfrac{3}{10}$, we get:

m = slope = $\dfrac{1}{2}$

c = y-intercept = $-\dfrac{3}{10}$

Hence, slope = $\dfrac{1}{2}, \text{y-intercept} = -\dfrac{3}{10}$.

(ii) Converting $\dfrac{x}{6} + \dfrac{y}{9} = 1$ in the form y = mx + c we get,

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

The equation of straight line is given by,

y = mx + c, where m is the slope and c is the y-intercept.

Comparing y = mx + c with $y = -\dfrac{3x}{2} + 9$, we get:

m = slope = $-\dfrac{3}{2}$

c = y-intercept = 9

Hence, slope = $-\dfrac{3}{2}$, y-intercept = 9.

(iii) Given,

x + 4 = 0

x = -4

This is a vertical line parallel to the y-axis, passing through the x-axis at x = -4.

We know that the inclination of a line parallel to y-axis is 90°.

∴ Slope of y-axis = tan 90° = infinity, which is not defined.

Since the line is parallel to the y-axis and passes through a negative x-value, it never crosses the y-axis. There is no y-intercept.

Hence, slope is not defined and line has no y-intercept.

(iv) Given,

y = 6

This is a horizontal line parallel to the x-axis, passing through the y-axis at y = 6.

We know that the inclination of a line parallel to x-axis is 0°.

∴ Slope of a line parallel to x-axis = tan 0° = 0.

y-intercept = 6

Hence, slope = 0 and y-intercept = 6.

Find the gradient and the equation of the line passing through the points:

(i) A(–2, 1) and B(3, –4)

(ii) A(4, –2) and B(2, –3)

Answer

(i) Slope of AB = $\dfrac{y_2 - y_1}{x_2 - x_1}$

= $\dfrac{-4 - 1}{3 - (-2)}$

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

= -1.

Equation : y - y₁ = m(x - x₁)

⇒ y - 1 = -1(x - (-2))

⇒ y - 1 = -1(x + 2)

⇒ y - 1 = -x - 2

⇒ x + y - 1 + 2 = 0

⇒ x + y + 1 = 0.

Hence, slope = -1, equation of AB is x + y + 1 = 0.

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

= $\dfrac{-3 - (-2)}{2 - 4}$

= $\dfrac{-1}{-2}$

= $\dfrac{1}{2}$.

Equation : y - y₁ = m(x - x₁)

⇒ y - (-2) = $\dfrac{1}{2}$ (x - 4)

⇒ 2(y + 2) = (x - 4)

⇒ 2y + 4 = x - 4

⇒ x - 2y - 4 - 4 = 0

⇒ x - 2y - 8 = 0.

Hence, slope = $\dfrac{1}{2}$, equation of AB is x - 2y - 8 = 0.

In the given diagram, ABC is a triangle, where B(4, -4) and C(-4, -2). D is a point on AC.

(a) Write down the coordinates of A and D.

(b) Find the coordinates of the centroid of ∆ABC.

(c) If D divides AC in the ratio k : 1, find the value of k.

(d) Find the equation of the line BD.

Answer

(a) From graph,

Co-ordinates of A = (0, 6) and D = (-3, 0)

(b) By formula,

Co-ordinates of centroid = $\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

$= \Big(\dfrac{0 + 4 + (-4)}{3}, \dfrac{6 + (-4) + (-2)}{3}\Big) \\[1em] = \Big(\dfrac{0}{3}, \dfrac{0}{3}\Big) \\[1em] = (0, 0).$

Hence, co-ordinates of centroid of ∆ABC = (0, 0).

(c) By section-formula,

(x, y) = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Given,

D divides AC in the ratio k : 1.

$\therefore (-3, 0) = \Big(\dfrac{k \times -4 + 1 \times 0}{k + 1}, \dfrac{k \times -2 + 1 \times 6}{k + 1}\Big) \\[1em] \Rightarrow (-3, 0) = \Big(\dfrac{-4k + 0}{k + 1}, \dfrac{-2k + 6}{k + 1}\Big) \\[1em] \Rightarrow (-3, 0) = \Big(\dfrac{-4k}{k + 1}, \dfrac{-2k + 6}{k + 1}\Big) \\[1em] \Rightarrow -3 = -\dfrac{4k}{k + 1} \text{ and } 0 = \dfrac{-2k + 6}{k + 1} \\[1em] \Rightarrow -3(k + 1) = -4k \text{ and } 0 = -2k + 6 \\[1em] \Rightarrow 3(k + 1) = 4k \text{ and } 2k = 6 \\[1em] \Rightarrow 3k + 3 = 4k \text{ and } k = \dfrac{6}{2} \\[1em] \Rightarrow 4k - 3k = 3 \text{ and } k = 3 \\[1em] \Rightarrow k = 3.$

Hence, k = 3.

(d) By two point form,

Equation of line :

y - y₁ = $\dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Equation of BD :

⇒ y - (-4) = $\dfrac{0 - (-4)}{-3 - 4}(x - 4)$

⇒ y + 4 = $\dfrac{4}{-7}(x - 4)$

⇒ -7(y + 4) = 4(x - 4)

⇒ -7y - 28 = 4x - 16

⇒ 4x + 7y - 16 + 28 = 0

⇒ 4x + 7y + 12 = 0.

Hence, equation of BD is 4x + 7y + 12 = 0.

A straight line passes through the points P(–1, 4) and Q(5, –2). It intersects x-axis and y-axis at the points A and B respectively and M is the mid-point of AB. Find :

(i) the equation of the line

(ii) the co-ordinates of A and B

(iii) the co-ordinates of M

Answer

(i) Given points, P(-1, 4) and Q(5, -2)

$\Rightarrow \text{ Slope of PQ } = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{-2 - 4}{5 - (-1)} \\[1em] = \dfrac{-6}{6} = -1.$

By point-slope form,

Equation of the line PQ, y - y₁ = m(x - x₁)

⇒ y - 4 = -1[x - (-1)]

⇒ y - 4 = -1[x + 1]

⇒ y - 4 = -x - 1

⇒ x + y = -1 + 4

⇒ x + y - 3 = 0.

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

(ii) For point A (on x-axis), y = 0.

So, putting y = 0 in the equation of PQ, we have

⇒ x + 0 = 3

⇒ x = 3.

∴ A = (3, 0).

For point B (on y-axis), x = 0.

So, putting x = 0 in the equation of PQ, we have

⇒ 0 + y = 3

⇒ y = 3

∴ B = (0, 3).

Hence, co-ordinates of A = (3, 0) and B = (0, 3).

(iii) M is the mid-point of AB.

∴ M = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

= $\Big(\dfrac{3 + 0}{2}, \dfrac{0 + 3}{2}\Big)$

= $\Big(\dfrac{3}{2}, \dfrac{3}{2}\Big)$

Hence, mid-point of AB = $\Big(\dfrac{3}{2}, \dfrac{3}{2}\Big)$.

A(2, 3) and B(–2, 5) are two given points. Find :

(i) the gradient of AB

(ii) the equation of AB

(iii) the co-ordinates of the point, where AB intersects x-axis.

Answer

(i) Given points, A(2, 3) and B(–2, 5)

$\Rightarrow \text{ Slope of AB } = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{5 - 3}{-2 -2} \\[1em] = \dfrac{2}{-4} = -\dfrac{1}{2}.$

Hence, slope = $-\dfrac{1}{2}$.

(ii) By point-slope form,

Equation of the line AB, y - y₁ = m(x - x₁)

⇒ y - 3 = $-\dfrac{1}{2}$(x - 2)

⇒ 2(y - 3) = -1(x - 2)

⇒ 2y - 6 = -x + 2

⇒ x + 2y = 6 + 2

⇒ x + 2y = 8.

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

(iii) The line intersects the x-axis when y = 0. Substituting y = 0 into the equation of the line, x + 2y = 8, we get :

⇒ x + 2(0) = 8

⇒ x = 8

Hence, coordinates of the point where AB intersects the x-axis are (8, 0).

A straight line passes through the points A(2, –4) and B(5, –2). Find :

(i) the slope of the line AB

(ii) the equation of the line AB

(iii) the value of k, if AB passes through the point P(k + 3, k – 4)

Answer

(i) Given points, A(2, –4) and B(5, –2)

$\Rightarrow \text{ Slope of AB } = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{-2 - (-4)}{5 - 2} \\[1em] = \dfrac{2}{3}.$

Hence, slope = $\dfrac{2}{3}$.

(ii) By point-slope form,

Equation of the line AB, y - y₁ = m(x - x₁)

⇒ y - (-4) = $\dfrac{2}{3}$(x - 2)

⇒ 3(y + 4) = 2(x - 2)

⇒ 3y + 12 = 2x - 4

⇒ 3y - 2x + 12 + 4 = 0

⇒ 3y - 2x + 16 = 0

Hence, equation of line is 3y - 2x + 16 = 0.

(iii) Since the line AB passes through the point P(k + 3, k - 4), the coordinates of P must satisfy the equation of the line, x - 2y - 10 = 0.

⇒ 3(k - 4) - 2(k + 3) + 16 = 0

⇒ 3k - 12 - 2k - 6 + 16 = 0

⇒ (3k - 2k) + (-12 - 6 + 16) = 0

⇒ k - 2 = 0

⇒ k = 2.

Hence, k = 2.

If A(3, 4), B(7, –2) and C(–2, –1) are the vertices of a ΔABC, write down the equation of the median through the vertex C.

Answer

Let median through C be CX.

We know that, the median, CX through C will bisect the line AB.

By Mid-point formula,

Mid-point = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

The co-ordinates of point X are

$\Big(\dfrac{3 + 7}{2}, \dfrac{4 + (-2)}{2}\Big) = \Big(\dfrac{10}{2}, \dfrac{2}{2}\Big)$ = (5, 1).

By formula,

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

Substituting values we get,

Slope of CX = $\dfrac{1 - (-1)}{5 - (-2)} = \dfrac{2}{7}$.

Then, the required equation of the median CX is given by :

⇒ y - y₁ = m(x - x₁)

⇒ y - (-1) = $\dfrac{2}{7}$[x - (2)]

⇒ 7(y + 1) = 2(x + 2)

⇒ 7y + 7 = 2x + 4

⇒ 7y = 2x + 4 - 7

⇒ 2x - 7y - 3 = 0

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

A(2, 5), B(–1, 2) and C(5, 8) are the vertices of a ΔABC, M is a point on AB such that AM : MB = 1 : 2. Find the co-ordinates of M. Hence, find the equation of the line passing through the points C and M.

Answer

Given AM : MB = 1 : 2. By section-formula the coordinates of M are,

$\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \Big)$

Substituting values we get,

$M = \Big(\dfrac{1(-1) + 2(2)}{1 + 2}, \dfrac{1(2) + 2(5)}{1 + 2} \Big) \\[1em] = \Big(\dfrac{-1 + 4}{3}, \dfrac{2 + 10}{3} \Big) \\[1em] = \Big(\dfrac{3}{3}, \dfrac{12}{3} \Big) \\[1em] = (1, 4).$

Equation of line CM can be given by two-point formula i.e.,

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

Substituting values we get,

⇒ y − 8 = $\dfrac{4 - 8}{1 - 5}$(x − 5)

⇒ y − 8 = $\dfrac{-4}{-4}$(x − 5)

⇒ y − 8 = 1(x − 5)

⇒ y − 8 = x − 5

⇒ x − y − 5 + 8 = 0

⇒ x − y + 3 = 0.

Hence, the equation of CM is x - y + 3 = 0 and the coordinates of M are (1, 4).

The vertices of a ΔABC are A(2, –11), B(2, 13) and C(–12, 1). Find the equations of its sides.

Answer

Given,

Coordinates A(2, −11) and B(2, 13)

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

Substituting values we get,

Slope of AB = $\dfrac{13 - (-11)}{2 - 2} = \dfrac{13 + 11}{0} =\dfrac{24}{0}$

Slope is not defined.

The line AB is a vertical line parallel to the y-axis.

Points have the same x-coordinate, x = 2.

Equation of line AB: x = 2

Given, Points: B(2,13), C(−12,1)

By formula,

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

Substituting values we get,

Slope of BC = $\dfrac{1 - (13)}{-12 - 2} = \dfrac{-12}{-14} =\dfrac{6}{7}$

By point-slope form,

Equation of the line BC, y - y₁ = m(x - x₁)

⇒ y - 13 = $\dfrac{6}{7}$ (x - 2)

⇒ 7(y - 13) = 6(x - 2)

⇒ 7y - 91 = 6x - 12

⇒ 7y - 6x = -12 + 91

⇒ 7y - 6x = 79

Equation of BC: 7y - 6x = 79

Given, Points: C(−12,1), A(2,−11)

By formula,

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

Substituting values we get,

Slope of CA = $\dfrac{-11 - 1}{2 - (-12)} = \dfrac{-12}{14} =-\dfrac{6}{7}$

By point-slope form,

Equation of the line CA, y - y₁ = m(x - x₁)

⇒ y - 1 = $-\dfrac{6}{7}$ (x + 12)

⇒ 7(y - 1) = -6(x + 12)

⇒ 7y - 7 = -6x - 72

⇒ 7y + 6x - 7 + 72 = 0

⇒ 7y + 6x + 65 = 0

Equation of the line CA: 7y + 6x + 65 = 0

Hence, the equation of AB, BC and CA are x = 2, 7y - 6x = 79, 7y + 6x + 65 = 0 respectively.

ABC is a triangle whose vertices are A(1, –1), B(0, 4) and C(–6, 4). D is the mid-point of BC. Find the :

(i) co-ordinates of D

(ii) equation of the median AD

Answer

(i) By formula,

Mid-point (M) = = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Given,

D is the mid-point of BC.

∴ Co-ordinates of D

$= \Big(\dfrac{0 + (-6)}{2}, \dfrac{4 + 4}{2}\Big) \\[1em] = \Big(\dfrac{-6}{2}, \dfrac{8}{2}\Big) \\[1em] = (-3, 4).$

Hence, coordinates of D = (-3, 4).

(ii) Slope = $\dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{4 - (-1)}{(-3) - 1} = -\dfrac{5}{4}$

Equation of a line :

y - y₁ = m(x - x₁)

Substituting values we get :

Equation of AD :

⇒ y - (-1) = $-\dfrac{5}{4}$ (x - 1)

⇒ -4(y + 1) = 5(x - 1)

⇒ -4y - 4 = 5x - 5

⇒ 5x + 4y = -4 + 5

⇒ 5x + 4y - 1 = 0.

Hence, equation of median AD is 5x + 4y - 1 = 0.

Find the equation of a line passing through the point (2, 3) and intersecting the line 2x – 3y = 6 on the y-axis.

Answer

On the y-axis,

x = 0.

Substitute x = 0 into 2x − 3y = 6:

⇒ 2(0) − 3y = 6

⇒ −3y = 6

⇒ y = −2

So, the line 2x − 3y = 6 meets the y-axis at (0, -2).

Calculating slope for points (2, 3) and (0, -2).

By formula,

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

m = $\dfrac{3 -(-2)}{2 - 0} = \dfrac{5}{2}$

By two-point form :

Equation of a line :

y - y₁ = m(x - x₁)

Substituting values we get :

⇒ y - (-2) = $\dfrac{5}{2}$ (x - 0)

⇒ 2(y + 2) = 5x

⇒ 2y + 4 = 5x

⇒ 5x - 2y - 4 = 0

Hence, equation of line 5x - 2y = 4.

Find the equation of a line with x-intercept = 5 and passing through the point (4, –3).

Answer

When x-intercept = 5; corresponding point on the x-axis = (5, 0).

By formula,

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

Slope of the line through (5, 0) and (4, -3) = $\dfrac{-3 - 0}{4 - 5} = \dfrac{-3}{-1} = 3$

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = 3(x − 5)

⇒ y = 3x - 15

⇒ 3x - y = 15.

Hence, equation of line is 3x - y = 15.

Find the equations of the diagonals of a rectangle whose sides are: x = –1, x = 4, y = –1 and y = 2.

Answer

From figure,

The lines intersect at point I, J, K and L.

By two point formula,

Equation of line : $y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Equation of diagonal IK is

$\Rightarrow y - 2 = \dfrac{-1 - 2}{4 - (-1)}[x - (-1)] \\[1em] \Rightarrow y - 2 = -\dfrac{3}{5}[x + 1] \\[1em] \Rightarrow 5(y - 2) = -3[x + 1] \\[1em] \Rightarrow 5y - 10 = -3x -3 \\[1em] \Rightarrow 5y + 3x -10 + 3 = 0 \\[1em] \Rightarrow 3x + 5y - 7 = 0 \\[1em] \Rightarrow 3x + 5y = 7.$

Equation of diagonal LJ is

⇒ y - (-1) = $\dfrac{2 - (-1)}{4 - (-1)}$ [x - (-1)]

⇒ y - (-1) = $\dfrac{3}{5}$ [x - (-1)]

⇒ 5(y + 1) = 3[x + 1]

⇒ 5y + 5 = 3x + 3

⇒ 5y − 3x + 5 − 3 = 0

⇒ 5y - 3x + 2 = 0

⇒ 3x - 5y = 2.

Hence, equation of diagonals are 3x - 5y = 2 and 3x + 5y = 7.

Find the equation of the line passing through the point (3, 2) and making positive equal intercepts on axes. Find the length of each intercept.

Answer

Let the line containing the point (3, 2) passes through x-axis at A(x, 0) and y-axis at B(0, y).

Given, the intercepts made on both the axes are equal.

∴ x = y

Slope of the line

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{0 - y}{x - 0} \\[1em] = \dfrac{-y}{x} \\[1em] = \dfrac{-x}{x} \\[1em] = -1.$

Hence, the equation of the line will be

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = -1(x - 3)

⇒ y - 2 = -x + 3

⇒ y + x - 2 - 3 = 0

⇒ y = -x + 5.

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

c = 5.

Thus, y-intercept = 5.

∴ x-intercept = 5.

Hence, equation of line is x + y = 5 and length of x and y intercept is 5 units.

Find the equation of the line passing through the origin and the point of intersection of the lines 5x + 7y = 3 and 2x – 3y = 7.

Answer

⇒ 5x + 7y = 3 ….(1)

⇒ 2x - 3y = 7 ….(2)

Multiplying equation (1) by 3, we get :

⇒ 3(5x + 7y) = 3.3

⇒ 15x + 21y = 9 ….(3)

Multiplying equation (2) by 7, we get :

⇒ 7(2x - 3y) = 7.7

⇒ 14x - 21y = 49 ….(4)

Adding equations (3) and (4) we get,

⇒ 15x + 21y + 14x - 21y = 9 + 49

⇒ 29x = 58

⇒ x = $\dfrac{58}{29}$

⇒ x = 2.

Substituting x = 2 in (1), we get :

⇒ 5(2) + 7y = 3

⇒ 10 + 7y = 3

⇒ 7y = 3 - 10

⇒ 7y = -7

⇒ y = $\dfrac{-7}{7}$

⇒ y = -1.

Hence, the point of intersection of lines is (2, -1).

The equation of the line joining (2, -1) and (0, 0) will be given by two-point form i.e.,

$y - y _1 = \dfrac{y_2 - y_1}{x_2 - x_1} (x - x_1)$

Substituting values in above equation we get,

⇒ y - (-1) = $\dfrac{0 - (-1)}{0 - 2}$ (x - 2)

⇒ y + 1 = $\dfrac{1}{-2}$ (x - 2)

⇒ -2(y + 1) = (x - 2)

⇒ -2y - 2 = x - 2

⇒ x + 2y - 2 + 2 = 0

⇒ x + 2y = 0.

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

M and N are two points on the x-axis and y-axis respectively. P(3, 2) divides the line segment MN in the ratio 2 : 3. Find :

(i) the co-ordinates of M and N

(ii) slope of the line MN

Answer

(i) Let the coordinates of M and N be (x, 0) and (0, y).

By section formula the coordinates of P are,

$\Rightarrow \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] = \Big(\dfrac{2(0) + 3x}{2 + 3}, \dfrac{2y + 3(0)}{2 + 3}\Big) \\[1em] = \Big(\dfrac{3x}{5}, \dfrac{2y}{5}\Big)$

Given, P(3, 2). Comparing two values of P we get,

⇒ 3 = $\dfrac{3x}{5}$ and 2 = $\dfrac{2y}{5}$

⇒ 3x = 15 and 2y = 10

⇒ x = 5 and y = 5.

Hence, the coordinates of M and N are (5, 0) and (0, 5) respectively.

(ii) Slope of line MN can be given by $\dfrac{y_2 - y_1}{x_2 - x_1}$

Substituting value in above equation we get slope,

= $\dfrac{5 - 0}{0 - 5}$

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

= -1.

Hence, the slope of the line is -1.

In what ratio does the line x – 5y + 15 = 0 divide the join of A(2, 1) and B(–3, 6)? Also, find the co-ordinates of their point of intersection.

Answer

Let P divides line AB in the ratio m : n.

By section formula,

$\Rightarrow P(x, y) = \Big(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\Big) \\[1em] = \Big(\dfrac{m(-3) + n(2)}{m + n}, \dfrac{m(6) + n(1)}{m + n}\Big) \\[1em] = \Big(\dfrac{-3m + 2n}{m + n}, \dfrac{6m + n}{m + n}\Big).$

Since, point P lies on line x - 5y + 15 = 0, substituting values we get :

$\Rightarrow \dfrac{-3m + 2n}{m + n} - 5 \Big(\dfrac{6m + n}{m + n}\Big) + 15 = 0 \\[1em] \Rightarrow \dfrac{-3m + 2n - 5(6m + n) + 15(m + n)}{(m + n)} = 0 \\[1em] \Rightarrow -3m + 2n - 30m - 5n + 15m + 15n = 0 \\[1em] \Rightarrow -18m + 12n = 0 \\[1em] \Rightarrow 6(-3m + 2n) = 0 \\[1em] \Rightarrow -3m + 2n = 0 \\[1em] \Rightarrow 2n = 3m \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{2}{3}.$

By section formula,

$\Rightarrow (x, y) = \Big(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{2(-3) + 3(2)}{2 + 3}, \dfrac{2(6) + 3(1)}{2 + 3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{-6 + 6}{5}, \dfrac{12 + 3}{5}\Big) \\[1em] \Rightarrow (x, y) = 0, \dfrac{15}{5} \\[1em] \Rightarrow (x, y) = (0, 3).$

Hence, the line x – 5y + 15 = 0 divides AB in the ratio 2 : 3 and the co-ordinates of their point of intersection are (0, 3).

Find the equations of the medians of ΔABC whose vertices are A(–1, 2), B(2, 1) and C(0, 4). Hence, find the co-ordinates of the centroid of ΔABC.

Answer

By using Midpoint formula,

(x, y) = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

D (Midpoint of BC): B(2, 1)and C(0, 4)

D = $\Big(\dfrac{2 + 0}{2}, \dfrac{1 + 4}{2}\Big) = \Big(1, \dfrac{5}{2})$

E (Midpoint of AC): A(-1, 2) and C(0, 4)

E = $\Big(\dfrac{-1 + 0}{2}, \dfrac{2 + 4}{2}\Big) = \Big(-\dfrac{1}{2}, 3\Big)$

F (Midpoint of AB): A(-1, 2) and B(2, 1)

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

Median AD through A(-1, 2) and $D\Big(1, \dfrac{5}{2}\Big)$

By slope formula:

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

Substitute values we get:

$m_{AD} = \dfrac{\dfrac{5}{2} - 2}{1 - (-1)} = \dfrac{\dfrac{1}{2}}{2} = \dfrac{1}{4}$

The equation of the line will be given by two-point form i.e.,

y - y₁ = m(x - x₁)

Substituting values in above equation we get,

⇒ y - 2 = $\dfrac{1}{4}$ (x + 4)

⇒ 4(y - 2) = (x + 1)

⇒ 4y - 8 = x + 1

⇒ x - 4y + 9 = 0

Median BE through B(2, 1) and $E\Big(-\dfrac{1}{2}, 3\Big)$

$m_{BE} = \dfrac{3 - 1}{-\dfrac{1}{2} - 2} = \dfrac{2}{-\dfrac{5}{2}} = -\dfrac{4}{5}$

The equation of the line will be given by two-point form i.e.,

⇒ y - 1 = $-\dfrac{4}{5}$ (x - 2)

⇒ 5(y - 1) = 4(x - 2)

⇒ 5y - 5 = 4x - 8

⇒ 4x + 5y - 13 = 0

Median CF through C(0, 4) and $F\Big(\dfrac{1}{2}, \dfrac{3}{2}\Big)$

$m_{CF} = \dfrac{\dfrac{3}{2} - 4}{\dfrac{1}{2} - 0} = \dfrac{-\dfrac{5}{2}}{\dfrac{1}{2}} = -5$

Since C(0, 4) is the y-intercept, we use y = mx + c:

⇒ y = -5x + 4

⇒ 5x + y - 4 = 0

By using centroid formula,

$G = \Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

Using A(-1, 2), B(2, 1), and C(0, 4):

$\Rightarrow G = \Big(\dfrac{-1 + 2 + 0}{3}, \dfrac{2 + 1 + 4}{3}\Big) \\[1em] = \Big(\dfrac{-1 + 2 + 0}{3}, \dfrac{2 + 1 + 4}{3}\Big) \\[1em] = \Big(\dfrac{1}{3}, \dfrac{7}{3}\Big).$

Hence, equations of medians are x - 4y + 9 = 0, 4x + 5y - 13 = 0 and 5x + y - 4 = 0, coordinates of centroid $\Big(\dfrac{1}{3}, \dfrac{7}{3}\Big)$.

Find the coordinates of the centroid P of the △ ABC, whose vertices are A(-1, 3), B(3, -1) and C(0, 0). Hence, find the equation of a line passing through P and parallel to AB.

Answer

By formula,

Centroid of triangle = $\Big(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\Big)$

Substituting values we get :

$\Rightarrow \text{Centroid of △ ABC} = \Big(\dfrac{-1 + 3 + 0}{3}, \dfrac{3 + (-1) + 0}{3}\Big) \\[1em] \Rightarrow P = \Big(\dfrac{2}{3}, \dfrac{2}{3}\Big).$

By formula,

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

Substituting values we get :

$\text{Slope of AB} = \dfrac{-1 - 3}{3 - (-1)} \\[1em] = \dfrac{-4}{4} \\[1em] = -1.$

We know that,

Slope of parallel lines are equal.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get :

Equation of line passing through P and parallel to AB :

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

Hence, required equation is 3x + 3y = 4.

Three vertices of a parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3, 2). Find :

(i) the co-ordinates of the fourth vertex D

(ii) length of diagonal BD

(iii) equation of side AB of the parallelogram ABCD

Answer

The parallelogram ABCD is shown in the figure below:

(i) We know that the diagonals of a parallelogram bisect each other. Let (x, y) be the coordinates of D.

Mid-point of diagonal AC = $\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} = \dfrac{3 + 3}{2}, \dfrac{6 + 2}{2} = (3, 4)$

Mid-point of diagonal BD = $\dfrac{5 + x}{2}, \dfrac{10 + y}{2}$

These two should be same. On equating we get,

$\dfrac{5 + x}{2} = 3, \dfrac{10 + y}{2} = 4$

⇒ 5 + x = 6 and 10 + y = 8

⇒ x = 6 − 5 and y = 8 − 10

⇒ x = 1 and y = −2.

Hence, coordinates of D are (1, -2).

(ii) By distance formula the distance between B(5, 10) and D(1, -2) is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting values we get BD,

$= \sqrt{(1 - 5)^2 + (-2 - 10)^2} \\[1em] = \sqrt{(-4)^2 + (-12)^2} \\[1em] = \sqrt{16 + 144} \\[1em] = \sqrt{160} \\[1em] = 4\sqrt{10}$

Hence, the length of diagonal BD is $4\sqrt{10}$ units.

(iii) Equation of side AB can be given by two point formula i.e.,

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

Substituting values we get,

⇒ y − 6 = $\dfrac{10 - 6}{5 - 3}$ (x - 3)

⇒ y − 6 = $\dfrac{4}{2}$ (x - 3)

⇒ (y − 6) = 2(x - 3)

⇒ (y − 6) = 2x - 6

⇒ 2x - y -6 + 6 = 0

⇒ 2x - y = 0

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

In the given figure, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively. Find :

(i) the co-ordinates of A

(ii) the length of AB and AC

(iii) the ratio in which Q divides AC

(iv) the equation of the line AC

Answer

(i) From figure,

The co-ordinates of A = (4, 0).

(ii) By distance formula,

D = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting values we get,

$AB = \sqrt{(-2 - 4)^2 + (3 - 0)^2} \\[1em] = \sqrt{(-6)^2 + (3)^2} \\[1em] = \sqrt{36 + 9} \\[1em] = \sqrt{45} \\[1em] = 3\sqrt{5}. \\[1em] AC = \sqrt{(-2 - 4)^2 + (-4 - 0)^2} \\[1em] = \sqrt{(-6)^2 + (-4)^2} \\[1em] = \sqrt{36 + 16} \\[1em] = \sqrt{52} \\[1em] = 2\sqrt{13}.$

Hence, length of $AB = 3\sqrt{5}\text{ and } AC = 2\sqrt{13}$.

(iii) From figure,

Q lies on y-axis.

∴ x co-ordinate of Q = 0.

Let co-ordinate of Q are (0, a).

Let ratio in which Q divides AC be k : 1.

By section-formula,

$Q = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Comparing x-coordinate we get :

$\Rightarrow 0 = \dfrac{k \times -2 + 1 \times 4}{k + 1} \\[1em] \Rightarrow 0 = -2k + 4 \\[1em] \Rightarrow 2k = 4 \\[1em] \Rightarrow k = 2.$

k : 1 = 2 : 1.

Hence, Q divides AC in the ratio 2 : 1.

(iv) By formula,

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

Slope of AC = $\dfrac{-4 - 0}{-2 - 4} = \dfrac{-4}{-6} = \dfrac{2}{3}$

By point-slope form,

Equation of AC is :

⇒ y - y₁ = m (x - x₁)

⇒ y - 0 = $\dfrac{2}{3}$ (x - 4)

⇒ 3y = 2x - 8

⇒ 2x - 3y = 8

Hence, equation of AC is 2x - 3y = 8.

If A(2, –3), B(–5, 1), C(7, –1) and D(0, k) be four points such that AB is parallel to CD, find the value of k.

Answer

AB is parallel to line segment CD means they must have the same gradient

By using slope formula,

m = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Given, points A(2, –3), B(–5, 1)

Substituting values we get,

$m_{AB} = \dfrac{1 - (-3)}{-5 - 2} = \dfrac{1 + 3}{-7} = -\dfrac{4}{7}$

Given, points C(7, –1) and D(0, k)

Substituting values we get,

$m_{CD} = \dfrac{k - (-1)}{0 - 7} = \dfrac{k + 1}{-7}$

Equate the Gradients:

$\Rightarrow -\dfrac{4}{7} = \dfrac{k + 1}{-7} \\[1em] \Rightarrow -\dfrac{4}{7} \times -7 = k + 1 \\[1em] \Rightarrow k + 1 = 4 \\[1em] \Rightarrow k = 4 - 1 \\[1em] \Rightarrow k = 3.$

Hence, value of k = 3.

Show that the lines x + 2y – 5 = 0 and 2x + 4y + 9 = 0 are parallel.

Answer

Given,

⇒ x + 2y – 5 = 0

Converting x + 2y - 5 = 0 in the form y = mx + c we get,

⇒ 2y = -x + 5

⇒ y = $\dfrac{-x}{2} + \dfrac{5}{2}$

The equation of straight line is given by,

y = mx + c, where m is the slope and c is the y-intercept.

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

⇒ m₁ = $-\dfrac{1}{2}$

Given,

⇒ 2x + 4y + 9 = 0

Converting 2x + 4y + 9 = 0 in the form y = mx + c we get,

⇒ 4y = -2x - 9

⇒ y = $\dfrac{-2x}{4} - \dfrac{9}{4}$

⇒ y = $\dfrac{-1x}{2} - \dfrac{9}{4}$

Comparing y = mx + c with y = $\dfrac{-1x}{2} - \dfrac{9}{4}$, we get:

⇒ m₂ = $-\dfrac{1}{2}$

Since the gradient of the first line is equal to the gradient of the second line.

The lines are parallel to each other.

Hence, proved that lines are parallel.

Find the value of k for which the lines kx + 2y + 3 = 0 and 8x + ky – 1 = 0 are parallel.

Answer

Since, the lines are parallel they have same gradient.

Given, kx + 2y + 3 = 0

Converting kx + 2y + 3 = 0 in the form y = mx + c, we get :

⇒ 2y = -kx - 3

⇒ y = $-\dfrac{k}{2}x - \dfrac{3}{2}$

The equation of straight line is given by,

y = mx + c, where m is the slope and c is the y-intercept.

Comparing y = mx + c with y = $-\dfrac{k}{2}x - \dfrac{3}{2}$, we get :

⇒ m₁ = $-\dfrac{k}{2}$

Given,

8x + ky - 1 = 0

Converting 8x + ky - 1 = 0 in the form y = mx + c we get,

⇒ ky = -8x + 1

⇒ y = $\dfrac{-8x}{k} + \dfrac{1}{k}$

Comparing y = mx + c with y = $\dfrac{-8x}{k} + \dfrac{1}{k}$, we get :

⇒ m₂ = $-\dfrac{8}{k}$

Since, lines are parallel, equating the gradients :

$\Rightarrow -\dfrac{k}{2} = -\dfrac{8}{k} \\[1em] \Rightarrow k(k) = 8(2) \\[1em] \Rightarrow k^2 = 16 \\[1em] \Rightarrow k = \sqrt{16} \\[1em] \Rightarrow k = \pm 4.$

Hence, value of k = ± 4.

If the lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel, find the relation connecting a and b.

Answer

Converting 2x - by + 5 = 0 in the form y = mx + c we get,

⇒ 2x - by + 5 = 0

⇒ by = 2x + 5

⇒ y = $\dfrac{2}{b}x + \dfrac{5}{b}$

m₁ = $\dfrac{2}{b}$

Converting ax + 3y = 2 in the form y = mx + c we get,

⇒ ax + 3y = 2

⇒ 3y = -ax + 2

⇒ y = $-\dfrac{a}{3}x + \dfrac{2}{3}$

m₂ = $-\dfrac{a}{3}$

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

m₁ = m₂

⇒ $\dfrac{2}{b} = -\dfrac{a}{3}$

⇒ 3 × 2 = −a × b

⇒ 6 = −ab

⇒ ab + 6 = 0

⇒ ab = −6.

Hence, the relation connecting a and b is ab = -6.

Prove that the line through A(–2, 6) and B(4, 8) is perpendicular to the line through C(8, 12) and D(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,

= $\dfrac{8 - 6}{4 - (-2)} = \dfrac{2}{6} = \dfrac{1}{3}$

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

=$\dfrac{24 - 12}{4 - 8} = \dfrac{12}{-4} = -3$

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

Product of slopes = -1.

Hence, the lines are perpendicular to each other.

If A(2, –5), B(–2, 5), C(k, 3) and D(1, 1) be four points such that AB and CD are perpendicular to each other, find the value of k.

Answer

By using slope formula,

m = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Given, points A(2, –5), B(–2, 5)

Substituting values we get,

$m_{AB} = \dfrac{5 - (-5)}{-2 - 2} = \dfrac{5 + 5}{-4} = -\dfrac{10}{4} = -\dfrac{5}{2}$

Given, points C(k, 3) and D(1, 1)

Substituting values we get,

$m_{CD} = \dfrac{1 - 3}{1 - k} = \dfrac{-2}{1 - k}$

Since the lines are perpendicular the product of the gradients is equal to -1:

$\Rightarrow \Big(-\dfrac{5}{2}\Big) \times \Big(\dfrac{-2}{1 - k}\Big) = -1 \\[1em] \Rightarrow \Big(\dfrac{10}{2(1 - k)}\Big) = -1 \\[1em] \Rightarrow \Big(\dfrac{5}{1 - k}\Big) = -1 \\[1em] \Rightarrow 5 = -1(1 - k) \\[1em] \Rightarrow 5 = -1 + k \\[1em] \Rightarrow k = 5 + 1 \\[1em] \Rightarrow k = 6.$

Hence, value of k = 6.

Prove that the lines 2x + 3y + 8 = 0 and 27x – 18y + 10 = 0 are perpendicular to each other.

Answer

Converting 2x + 3y + 8 = 0 in the form y = mx + c we get,

⇒ 3y = -2x - 8

⇒ y = $-\dfrac{2}{3}x - \dfrac{8}{3}$

Comparing, we get slope of this line : m₁ = $-\dfrac{2}{3}$

Converting 27x – 18y + 10 = 0 in the form y = mx + c we get,

⇒ -18y = -27x - 10

⇒ y = $\dfrac{-27x}{-18} - \dfrac{10}{-18}$

⇒ y = $\dfrac{3x}{2} + \dfrac{5}{9}$

Comparing, we get slope of this line : m₂ = $\dfrac{3}{2}$

Product of Gradients

m₁ × m₂ = $-\dfrac{2}{3} \times \dfrac{3}{2}$

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

= -1

Since the product of the gradients is -1. The lines are perpendicular to each other.

Hence, proved that lines 2x + 3y + 8 = 0 and 27x – 18y + 10 = 0 are perpendicular to each other.

If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.

Answer

Given lines,

⇒ y = 3x + 7 and 2y + px = 3

⇒ y = 3x + 7 and 2y = -px + 3

⇒ y = 3x + 7 and y = $-\dfrac{p}{2}x + \dfrac{3}{2}$

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

Slope of 1st line = 3

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

Since,

Product of slopes of perpendicular lines = -1.

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

Hence, p = $\dfrac{2}{3}$.

If the straight lines 3x – 5y = 7 and 4x + ay + 9 = 0 are perpendicular to each other, 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,

$\Rightarrow m_1 \cdot m_2 = -1 \\[1em] \Rightarrow \dfrac{3}{5} \times -\dfrac{4}{a} = -1 \\[1em] \Rightarrow -\dfrac{12}{5a} = -1 \\[1em] \Rightarrow a = \dfrac{12}{5}.$

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

Without using Pythagoras Theorem, prove that the points A(1, 3), B(3, –1) and C(–5, –5) are the vertices of a right-angled triangle.

Answer

By using slope formula,

m = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Given, points A(1, 3), B(3, –1)

Substituting values we get,

$m_{AB} = \dfrac{-1 - 3}{3 - 1} = -\dfrac{4}{2} = -2$

Given, points B(3, –1) and C(–5, –5)

Substituting values we get,

$m_{BC} = \dfrac{-5 - (-1)}{-5 - 3} = \dfrac{-5 + 1}{-8} = \dfrac{-4}{-8} = \dfrac{1}{2}$

Check for perpendicularity,

$\Rightarrow m_{AB} \times m_{BC} = -2 \times \dfrac{1}{2} = -1.$

Since the product of the gradients of AB and BC is -1, the side AB is perpendicular to the side BC.

∠ABC = 90°.

Hence, proved the points A(1, 3), B(3, –1) and C(–5, –5) are the vertices of a right-angled triangle.

Without using distance formula, show that the points A(1, –2), B(3, 6), C(5, 10) and D(3, 2) are the vertices of a parallelogram.

Answer

By using slope formula,

m = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Given, points A(1, –2), B(3, 6)

Substituting values we get,

$m_{AB} = \dfrac{6 - (-2)}{3 - 1} = \dfrac{8}{2} = 4$

Given, points C(5, 10) and D(3, 2)

Substituting values we get,

$m_{DC} = \dfrac{10 - 2}{5 - 3} = \dfrac{8}{2} = 4$

Given, points B(3, 6), C(5, 10)

Substituting values we get,

$m_{BC} = \dfrac{10 - 6}{5 - 3} = \dfrac{4}{2} = 2$

Given, points A(1, –2), D(3, 2)

Substituting values we get,

$m_{AD} = \dfrac{2 - (-2)}{3 - 1} = \dfrac{4}{2} = 2$

m_AB = m_CD and m_BC = m_AD

∴ AB is parallel to CD and BC is parallel to AD

Since both pairs of opposite sides are parallel, the quadrilateral ABCD is a parallelogram.

Hence, proved that ABCD is a parallelogram.

Given that A(5, 4), B(–3, –2) and C(1, –8) are the vertices of a ΔABC. Find:

(i) the slope of median AD

(ii) the slope of altitude BM

Answer

(i) Since, AD is median. So, D is the mid-point of BC.

By using formula,

(x, y) = $\Big(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\Big)$

Substitute values we get,

D = $\Big(\dfrac{-3 + 1}{2}, \dfrac{-2 + (-8)}{2}\Big) = \Big(\dfrac{-2}{2}, \dfrac{-10}{2}\Big) = (-1, -5)$

By using slope formula,

m = $\dfrac{y_2 - y_1}{x_2 - x_1}$

Slope of AD = $\dfrac{-5 - 4}{-1 - 5} = \dfrac{-9}{-6} = \dfrac{3}{2}$

Hence, slope of the median AD = $\dfrac{3}{2}$.

(ii) The altitude BM is perpendicular to the side AC. Therefore, the product of their slopes is -1.

Slope of AC = $\dfrac{-8 - 4}{1 - 5} = \dfrac{-12}{-4} = 3$

m_BM × 3 = -1

m_BM = $\dfrac{-1}{3}$

Hence, slope of the BM = $-\dfrac{1}{3}$.