Section & Mid-Point Formula — Exercise 13(B)

Point A(3, 4) is the center of a circle. If one of its diameters has one end as (7, 8); the other end of this diameter is :

1. (1, 0)

2. (0, 1)

3. (-1, 0)

4. (0, -1)

Answer

Given,

A(3, 4) is center and let B(7, 8) be one end of the diameter.

Let other end of diameter be C(x, y).

By mid-point formula,

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

From figure,

A is the mid-point of BC.

$\therefore (3, 4) = \Big(\dfrac{x + 7}{2}, \dfrac{y + 8}{2}\Big) \\[1em] \Rightarrow \dfrac{x + 7}{2} = 3 \text{ and } \dfrac{y + 8}{2} = 4 \\[1em] \Rightarrow x + 7 = 6 \text{ and } y + 8 = 8 \\[1em] \Rightarrow x = 6 - 7 \text{ and } y = 8 - 8 \\[1em] \Rightarrow x = -1 \text{ and } y = 0.$

∴ C = (-1, 0).

Hence, Option 3 is the correct option.

Point A lies on x-axis and point B lies on y-axis. If P(2, -2) bisects the line segment AB, the co-ordinates of A are :

1. (4, 0)

2. (0, 4)

3. (-4, 0)

4. (0, -4)

Answer

Given,

Point A lies on x-axis.

∴ Let point A be (x, 0).

Point B lies on y-axis.

∴ Let point B be (0, y).

P(2, -2) bisects the line segment AB or P is the mid-point of AB.

By mid-point formula,

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

Substituting values we get :

$\Rightarrow (2, -2) = \Big(\dfrac{x + 0}{2}, \dfrac{0 + y}{2}\Big) \\[1em] \Rightarrow (2, -2) = \Big(\dfrac{x}{2}, \dfrac{y}{2}\Big) \\[1em] \Rightarrow \dfrac{x}{2} = 2 \text{ and } \dfrac{y}{2} = -2 \\[1em] \Rightarrow x = 4 \text{ and } y = -4.$

Co-ordinates of A = (x, 0) = (4, 0).

Hence, Option 1 is the correct option.

In parallelogram ABCD, A = (6, 0), B = (12, -4) and C = (4, -4); then the co-ordinates of vertex D are :

1. (2, 0)

2. (-2, 0)

3. (0, 2)

4. (0, -2)

Answer

Let co-ordinates of vertex D be (x, y).

We know that,

Diagonal of parallelogram bisect each other.

From figure,

O (a, b) is the mid-point of AC.

By mid-point formula,

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

Substituting values we get :

$\Rightarrow (a, b) = \Big(\dfrac{6 + 4}{2}, \dfrac{0 + (-4)}{2}\Big) \\[1em] = \Big(\dfrac{10}{2}, \dfrac{-4}{2}\Big) \\[1em] = (5, -2).$

From figure,

O is also the mid-point of BD.

$\therefore (5, -2) = \Big(\dfrac{12 + x}{2}, \dfrac{-4 + y}{2}\Big) \\[1em] \Rightarrow \dfrac{12 + x}{2} = 5 \text{ and } \dfrac{-4 + y}{2} = -2 \\[1em] \Rightarrow 12 + x = 10 \text{ and } -4 + y = -4 \\[1em] \Rightarrow x = 10 - 12 \text{ and } y = -4 + 4 \\[1em] \Rightarrow x = -2 \text{ and } y = 0.$

D = (-2, 0).

Hence, Option 2 is the correct option.

The point P(2, -7) is reflected in the point (0, 3); the co-ordinates of the image of point P are :

1. (2, 13)

2. (2, -13)

3. (-2, -13)

4. (-2, 13)

Answer

Let point P on reflection in the point (0, 3) becomes P'(x, y).

So, (0, 3) will be the mid-point of PP'.

By formula,

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

Substituting values we get :

$\Rightarrow (0, 3) = \Big(\dfrac{2 + x}{2}, \dfrac{-7 + y}{2}\Big) \\[1em] \Rightarrow 0 = \dfrac{2 + x}{2} \text{ and } 3 = \dfrac{-7 + y}{2} \\[1em] \Rightarrow 2 + x = 0 \text{ and } -7 + y = 6 \\[1em] \Rightarrow x = -2 \text{ and } y = 6 + 7 \\[1em] \Rightarrow x = -2 \text{ and } y = 13.$

P' = (-2, 13).

Hence, Option 4 is the correct option.

The co-ordinates of the centroid of a triangle with vertices (-6, -3), (0, 0) and (12, -6) are :

1. (2, 3)

2. (-2, 3)

3. (2, -3)

4. (-2, -3)

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 :

$\text{Centroid } = \Big(\dfrac{-6 + 0 + 12}{3}, \dfrac{-3 + 0 + (-6)}{3}\Big) \\[1em] = \Big(\dfrac{6}{3}, \dfrac{-9}{3}\Big) \\[1em] = (2, -3).$

Hence, Option 3 is the correct option.

Points A and B have co-ordinates (3, 5) and (x, y) respectively. The mid-point of AB is (2, 3). Find the values of x and y.

Answer

By formula,

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

Given, M = (2, 3)

Substituting values in above formula,

$(2, 3) = \Big(\dfrac{3 + x}{2}, \dfrac{5 + y}{2}\Big) \\[1em] \therefore 2 = \dfrac{3 + x}{2} \text{ and } 3 = \dfrac{5 + y}{2} \\[1em] \Rightarrow 4 = 3 + x \text{ and } 6 = 5 + y \\[1em] \Rightarrow x = 1 \text{ and } y = 1.$

Hence, x = 1 and y = 1.

Given M is the mid-point of AB, find the co-ordinates of:

(i) A; if M = (1, 7) and B = (-5, 10),

(ii) B; if A = (3, -1) and M = (-1, 3).

Answer

By formula,

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

(i) Let co-ordinates of A = (x, y).

Substituting values in above formula we get,

$(1, 7) = \Big[\dfrac{x + (-5)}{2}, \dfrac{y + 10}{2}\Big] \\[1em] \therefore 1 = \dfrac{x - 5}{2} \text{ and } 7 = \dfrac{y + 10}{2} \\[1em] \Rightarrow 2 = x - 5 \text{ and } 14 = y + 10 \\[1em] \Rightarrow x = 7 \text{ and } y = 4.$

A = (x, y) = (7, 4).

Hence, co-ordinates of A = (7, 4).

(ii) Let co-ordinates of B = (x, y).

Substituting values in above formula we get,

$(-1, 3) = \Big(\dfrac{3 + x}{2}, \dfrac{-1 + y}{2}\Big) \\[1em] \therefore -1 = \dfrac{3 + x}{2} \text{ and } 3 = \dfrac{y - 1}{2} \\[1em] \Rightarrow -2 = x + 3 \text{ and } 6 = y - 1 \\[1em] \Rightarrow x = -5 \text{ and } y = 7.$

B = (x, y) = (-5, 7).

Hence, co-ordinates of B = (-5, 7).

P(-3, 2) is the mid-point of line segment AB as shown in the given figure. Find the co-ordinates of points A and B.

Answer

Since, A lie on y-axis, let its co-ordinates be (0, y) and B lie on x-axis, let its co-ordinates be (x, 0).

P is mid-point of AB.

By formula,

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

Substituting values we get,

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

A = (0, y) = (0, 4) and B = (x, 0) = (-6, 0).

Hence, the co-ordinates of points A and B are (0, 4) and (-6, 0).

(-5, 2), (3, -6) and (7, 4) are the vertices of a triangle. Find the length of its median through the vertex (3, -6).

Answer

Let A = (3, -6), B = (-5, 2) and C = (7, 4).

From figure, AD is the median.

Since, AD is median so, BD = DC.

Thus, D is mid-point of BC.

By formula,

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

Substituting value we get,

$D = \Big(\dfrac{-5 + 7}{2}, \dfrac{2 + 4}{2}\Big) \\[1em] = \Big(\dfrac{2}{2}, \dfrac{6}{2}\Big) \\[1em] = (1, 3).$

Distance between two points = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting values we get,

$AD = \sqrt{(1 - 3)^2 + [3 - (-6)]^2} \\[1em] = \sqrt{(-2)^2 + (9)^2} \\[1em] = \sqrt{4 + 81} \\[1em] = \sqrt{85} \\[1em] = 9.22$

Hence, the length of its median through the vertex (3, -6) = 9.22 units.

Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.

Answer

Let co-ordinates of A be (x, y) and D be (p, q).

Since, AB = BC.

B is the mid-point of AC.

By formula,

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

Substituting values we get,

$\Rightarrow B = \Big(\dfrac{x + 1}{2}, \dfrac{y + 8}{2}\Big) \\[1em] \Rightarrow (0, 3) = \Big(\dfrac{x + 1}{2}, \dfrac{y + 8}{2}\Big) \\[1em] \Rightarrow 0 = \dfrac{x + 1}{2} \text{ and } 3 = \dfrac{y + 8}{2} \\[1em] \Rightarrow x + 1 = 0 \text{ and } y + 8 = 6 \\[1em] \Rightarrow x = -1 \text{ and } y = -2.$

A = (x, y) = (-1, -2).

Since, BC = CD.

C is mid-point of BD.

By formula,

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

Substituting values we get,

$\Rightarrow C = \Big(\dfrac{0 + p}{2}, \dfrac{3 + q}{2}\Big) \\[1em] \Rightarrow (1, 8) = \Big(\dfrac{p}{2}, \dfrac{3 + q}{2}\Big) \\[1em] \Rightarrow 1 = \dfrac{p}{2} \text{ and } 8 = \dfrac{3 + q}{2} \\[1em] \Rightarrow p = 2 \text{ and } 3 + q = 16 \\[1em] \Rightarrow p = 2 \text{ and } q = 13.$

D = (p, q) = (2, 13).

Hence, the co-ordinates of A = (-1, -2) and D = (2, 13).

A (2, 5), B (1, 0), C(-4, 3) and D (-3, 8) are the vertices of quadrilateral ABCD. Find the co-ordinates of the mid-points of AC and BD. Give a special name to the quadrilateral.

Answer

By formula,

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

Let mid-point of AC be E.

Substituting value we get,

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

Let mid-point of BD be F.

Substituting value we get,

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

Thus, the co-ordinates of the mid-points of AC and BD are same i.e., AC and BD bisect each other.

∴ ABCD is a parallelogram.

Hence, the co-ordinates of the mid-points of AC = (-1, 4) and BD = (-1, 4) and ABCD is a parallelogram.

P (4, 2) and Q (-1, 5) are the vertices of parallelogram PQRS and (-3, 2) are the co-ordinates of the point of intersection of the diagonals. Find co-ordinates of R and S.

Answer

We know that diagonals of a parallelogram bisect each other.

Let co-ordinates of R= (a, b) and S = (c, d).

From figure,

O is the mid-point of PR.

By formula,

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

Substituting value we get,

$\Rightarrow (-3, 2) = \Big(\dfrac{4 + a}{2}, \dfrac{2 + b}{2}\Big) \\[1em] \Rightarrow -3 = \dfrac{4 + a}{2} \text{ and } 2 = \dfrac{2 + b}{2} \\[1em] \Rightarrow 4 + a = -6 \text{ and } 2 + b = 4 \\[1em] \Rightarrow a = -10 \text{ and } b = 2.$

R = (a, b) = (-10, 2).

O is also the mid-point of QS,

$\Rightarrow (-3, 2) = \Big(\dfrac{-1 + c}{2}, \dfrac{5 + d}{2}\Big) \\[1em] \Rightarrow -3 = \dfrac{-1 + c}{2} \text{ and } 2 = \dfrac{5 + d}{2} \\[1em] \Rightarrow -1 + c = -6 \text{ and } 5 + d = 4 \\[1em] \Rightarrow c = -5 \text{ and } d = -1.$

S = (c, d) = (-5, -1).

Hence, the co-ordinates of R = (-10, 2) and S = (-5, -1).

The points (2, -1), (-1, 4) and (-2, 2) are mid-points of the sides of a triangle. Find its vertices.

Answer

Let D = (2, -1), E = (-1, 4) and F = (-2, 2).

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of △ABC.

Mid-point of AB, i.e. D(2, -1).

$\therefore 2 = \dfrac{x_1 + x_2}{2} \text{ and } -1 = \dfrac{y_1 + y_2}{2}$

⇒ x₁ + x₂ = 4 ........(1)

⇒ y₁ + y₂ = -2 ........(2)

Mid-point of BC, i.e. E(-1, 4).

$\therefore -1 = \dfrac{x_2 + x_3}{2} \text{ and } 4 = \dfrac{y_2 + y_3}{2}$

⇒ x₂ + x₃ = -2 ........(3)

⇒ y₂ + y₃ = 8 ........(4)

Mid-point of AC, i.e. F(-2, 2).

$\therefore -2 = \dfrac{x_1 + x_3}{2} \text{ and } 2 = \dfrac{y_1 + y_3}{2}$

⇒ x₁ + x₃ = -4 ........(5)

⇒ y₁ + y₃ = 4 ........(6)

Adding 1, 3 and 5 we get,

⇒ x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = 4 + (-2) + (-4)

⇒ 2(x₁ + x₂ + x₃) = -2

⇒ x₁ + x₂ + x₃ = -1.

From (1),

⇒ 4 + x₃ = -1

⇒ x₃ = -5.

Substituting value of x₃ in (5) we get,

⇒ x₁ + (-5) = -4

⇒ x₁ = 1.

Substituting value of x₃ in (3) we get,

⇒ x₂ + (-5) = -2

⇒ x₂ = 3.

Adding (2), (4) and (6) we get,

⇒ y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = -2 + 8 + 4

⇒ 2(y₁ + y₂ + y₃) = 10

⇒ y₁ + y₂ + y₃ = 5

From (2)

⇒ -2 + y₃ = 5

⇒ y₃ = 7.

Substituting value of y₃ in (4) we get,

⇒ y₂ + 7 = 8

⇒ y₂ = 1.

Substituting value of y₃ in (6) we get,

⇒ y₁ + 7 = 4

⇒ y₁ = -3.

A = (x₁, y₁) = (1, -3), B = (x₂, y₂) = (3, 1), C = (x₃, y₃) = (-5, 7).

Hence, A = (1, -3), B = (3, 1) and C = (-5, 7).

Points A(-5, x), B(y, 7) and C(1, -3) are collinear (i.e. lie on same straight line) such that AB = BC. Calculate the values of x and y.

Answer

Since, A, B and C are collinear and AB = BC.

We can say that B is the mid-point of AC.

By formula,

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

Substituting value we get,

$\Rightarrow B = \Big(\dfrac{-5 + 1}{2}, \dfrac{x + (-3)}{2}\Big) \\[1em] \Rightarrow (y, 7) = \Big(\dfrac{-4}{2}, \dfrac{x - 3}{2}\Big) \\[1em] \Rightarrow y = \dfrac{-4}{2} \text{ and } 7 = \dfrac{x - 3}{2} \\[1em] \Rightarrow y = -2 \text{ and } x - 3 = 14 \\[1em] \Rightarrow y = -2 \text{ and } x = 17.$

Hence, x = 17 and y = -2.

Points P(a, -4), Q(-2, b) and R(0, 2) are collinear. If Q lies between P and R, such that PR = 2QR, calculate the values of a and b.

Answer

From figure,

PR = PQ + QR

Given,

⇒ PR = 2QR

⇒ PQ +QR = 2QR

⇒ PQ = QR.

∴ Q is the mid-point of P and R.

By formula,

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

Substituting value we get,

$\Rightarrow Q = \Big(\dfrac{a + 0}{2}, \dfrac{-4 + 2}{2}\Big) \\[1em] \Rightarrow (-2, b) = \Big(\dfrac{a}{2}, \dfrac{-2}{2}\Big) \\[1em] \Rightarrow -2 = \dfrac{a}{2} \text{ and } b = \dfrac{-2}{2} \\[1em] \Rightarrow a = -4 \text{ and } b = -1.$

Hence, a = -4 and b = -1.

The co-ordinates of the centroid of a triangle PQR are (2, -5). If Q = (-6, 5) and R = (11, 8); calculate the co-ordinates of vertex P.

Answer

Let co-ordinates of P = (x, y).

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

Substituting values we get,

$\Rightarrow (2, -5) = \Big(\dfrac{x + (-6) + 11}{3}, \dfrac{y + 5 + 8}{3}\Big) \\[1em] \Rightarrow 2 = \dfrac{x - 6 + 11}{3} \text{ and } -5 = \dfrac{y + 13}{3} \\[1em] \Rightarrow 6 = x + 5 \text{ and } -15 = y + 13 \\[1em] \Rightarrow x = 1 \text{ and } y = -28.$

P = (x, y) = (1, -28).

Hence, co-ordinates of P = (1, -28).

A (5, x), B(-4, 3) and C (y, -2) are the vertices of the triangle ABC whose centroid is the origin. Calculate the values of x and y.

Answer

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

Substituting values we get,

$\Rightarrow (0, 0) = \Big[\dfrac{5 + (-4) + y}{3}, \dfrac{x + 3 + (-2)}{3}\Big] \\[1em] \Rightarrow 0 = \dfrac{5 - 4 + y}{3} \text{ and } 0 = \dfrac{x + 3 - 2}{3} \\[1em] \Rightarrow 0 = y + 1 \text{ and } 0 = x + 1 \\[1em] \Rightarrow y = -1 \text{ and } x = -1.$

Hence, x = -1 and y = -1.