Section Formula — Multiple Choice Questions

The points A(9, 0), B(9, 6), C(-9, 6) and D(-9, 0) are the vertices of a

1. rectangle

2. square

3. rhombus

4. trapezium

Answer

We know that,

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

Given, A(9, 0), B(9, 6), C(-9, 6) and D(-9, 0)

Calculating the length of sides by distance formula.

$AB = \sqrt{(9 - 9)^2 + (6 - 0)^2} \\[1em] = \sqrt{0 + 36} \\[1em] = 6 \text{ units.} \\[1em] BC = \sqrt{(-9 - 9)^2 + (6 - 6)^2} \\[1em] = \sqrt{(-18)^2 + 0^2} \\[1em] = \sqrt{324} \\[1em] = 18 \text{ units.} \\[1em] CD = \sqrt{(-9 - (-9))^2 + (0 - 6)^2} \\[1em] = \sqrt{0 + 36} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ units.} \\[1em] DA = \sqrt{(9 - (-9))^2 + (0 - 0)^2} \\[1em] = \sqrt{18^2 + 0} \\[1em] = \sqrt{324} \\[1em] = 18 \text{ units.}$

Since, AB = CD and BC = DA means opposite sides are equal.

Hence, Option 1 is the correct option.

If $P(\dfrac{a}{3}, 4)$ is the mid-point of the line segment joining the points Q(-6, 5) and R(-2, 3), then the value of a is

1. -4

2. -6

3. 12

4. -12

Answer

We know that,

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

Given, P is the midpoint of line segment QR. Putting the values in mid-point formula for x-coordinate we get,

$\Rightarrow \dfrac{a}{3} = \dfrac{-6 + (-2)}{2} \\[1em] \Rightarrow a = \dfrac{-8}{2} \times 3 \\[1em] \Rightarrow a = -4 \times 3 \\[1em] \Rightarrow a = -12.$

Hence, Option 4 is the correct option.

If the end points of a diameter of a circle are A(-2, 3) and B(4, -5) then the coordinates of its centre are

1. (2, -2)

2. (1, -1)

3. (-1, 1)

4. (-2, 2)

Answer

We know that,

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

Given A and B are end points of the diameter so by mid-point formula the coordinates of the centre of the circle are,

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

Hence, Option 2 is the correct option.

If one end of a diameter of a circle is (2,3) and the centre is (-2, 5), then the other end is

1. (-6, 7)

2. (6, -7)

3. (0, 8)

4. (0, 4)

Answer

Let B(x, y) be the other end of the diameter, whose one end is A(2, 3)

∴ The mid-point of AB is $\Big(\dfrac{x + 2}{2}, \dfrac{y + 3}{2}\Big)$

The center of the circle is (-2, 5).

Since, the centre of the circle is the mid-point of AB,

$\Rightarrow \dfrac{2 + x}{2} = -2 \text{ and } \dfrac{y + 3}{2} = 5 \\[1em] \Rightarrow 2 + x = -4 \text{ and } y + 3 = 10 \\[1em] \Rightarrow x = -4 - 2 \text{ and } y = 10 - 3 \\[1em] \Rightarrow x = -6 \text{ and } y = 7.$

∴ Coordinates of other end are (-6, 7).

Hence, Option 1 is the correct option.

If the mid-point of the line segment joining the points P(a, b - 2) and Q(-2, 4) is R(2, -3), then the values of a and b are

1. a = 4, b = -5

2. a = 6, b = 8

3. a = 6, b = -8

4. a = -6, b = 8

Answer

The mid-point of the line segment joining the points P(a, b - 2) and Q(-2, 4) is $\dfrac{a + (-2)}{2}, \dfrac{b - 2 + 4}{2}$.

Given R(2, -3) is the mid-point.

Comparing the values of mid-point we get,

$\dfrac{a + (-2)}{2} = 2 \text{ and } \dfrac{b - 2 + 4}{2} = -3$
⇒ a - 2 = 4 and b + 2 = -6
⇒ a = 4 + 2 and b = -6 - 2
⇒ a = 6 and b = -8.

Hence, Option 3 is the correct option.

The point which lies on the perpendicular bisector of the line segment joining the points A(-2, -5) and B(2, 5) is

1. (0, 0)

2. (0, 2)

3. (2, 0)

4. (-2, 0)

Answer

The point which lies on the perpendicular bisector of the line segment is the mid-point of the line.

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

Putting values to find mid point of AB,

$= \Big(\dfrac{-2 + 2}{2}, \dfrac{-5 + 5}{2}\Big) = (0, 0).$

Hence, Option 1 is the correct option.

The coordinates of the point which is equidistant from the three vertices of △AOB (shown in the adjoining figure) are

1. (x, y)

2. (y, x)

3. $\Big(\dfrac{x}{2}, \dfrac{y}{2}\Big)$

4. $\Big(\dfrac{y}{2}, \dfrac{x}{2}\Big)$

Answer

Since the triangle AOB is the right angled triangle the point which is equidistant from O, A and B is the mid-point of AB. Let that mid-point be D.

Applying mid-point formula, coordinates are,

$= \dfrac{0 + 2x}{2}, \dfrac{2y + 0}{2} \\[1em] = \dfrac{2x}{2}, \dfrac{2y}{2} \\[1em] = (x, y).$

Hence, Option 1 is the correct option.

The fourth vertex D of a parallelogram ABCD whose vertices are A(-2, 3), B(6, 7) and C(8, 3) is

1. (0, 1)

2. (0, -1)

3. (-1, 0)

4. (1, 0)

Answer

ABCD is a parallelogram whose three vertices are A(-2, 3), B(6, 7) and C(8, 3). Let coordinates of its fourth vertex D be (x, y).

The diagonals AC and BD bisect each other at O so O is the mid-point of AC as well as BD.

So, by mid-point formula the coordinates of O are,

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

Since, (3, 3) is the mid-point of BD so,

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

∴ Coordinates of D are (0, -1).

Hence, Option 2 is the correct option.

The point which divides the line segment joining the points (7, -6) and (3, 4) in the ratio 1 : 2 internally lies in the

1. Ist quadrant

2. IInd quadrant

3. IIIrd quadrant

4. IVth quadrant

Answer

Let P(x, y) be the point which divides the line segment joining the points.

Given, A(7, -6) and B(3, 4) is divided by P in ratio 1 : 2.

By section formula,

$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \text{ and } y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] = \dfrac{1 \times 3 + 2 \times 7}{1 + 2} \text{ and } \dfrac{1 \times 4 + 2 \times (-6)}{1 + 2} \\[1em] = \dfrac{3 + 14}{3} \text{ and } \dfrac{4 - 12}{3} \\[1em] = \dfrac{17}{3} \text{ and } -\dfrac{8}{3}.$

We see that x is positive and y is negative.

∴ It lies in the fourth quadrant.

Hence, Option 4 is the correct option.

The centroid of the triangle whose vertices are (-4, -2), (6, 2) and (4, 6) is

1. (2, 2)

2. (2, 3)

3. (3, 3)

4. (0, -1)

Answer

By formula,

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

Centroid of triangle ABC (G)

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

Hence, Option 1 is the correct option.

A(1, 4), B(4, 1) and C(x, 4) are the vertices of △ABC. If the centroid of the triangles is G(4, 3), then x is equal to

1. 2

2. 1

3. 7

4. 4

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 :

$\Rightarrow (4, 3) = \Big(\dfrac{1 + 4 + x}{3}, \dfrac{4 + 1 + 4}{3}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\dfrac{x + 5}{3}, \dfrac{9}{3}\Big) \\[1em] \Rightarrow (4, 3) = \Big(\dfrac{x + 5}{3}, 3\Big) \\[1em] \Rightarrow 4 = \dfrac{x + 5}{3} \\[1em] \Rightarrow x + 5 = 12 \\[1em] \Rightarrow x = 12 - 5 = 7.$

Hence, option 3 is the correct option.