Section & Mid-Point Formulae — Multiple Choice Questions

The coordinates of the point P which divides the join of A(5, -2) and B(9, 6) in the ratio 3 : 1 are :

1. (4, -7)

2. $\Big(\dfrac{7}{2}, 4\Big)$

3. (8, 4)

4. (12, 8)

Answer

Let point P be (x, y).

Given,

m₁ : m₂ = 3 : 1

By section-formula,

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

Substituting values we get :

$\Rightarrow (x, y) = \Big(\dfrac{3 \times 9 + 1 \times 5}{3 + 1}, \dfrac{3 \times 6 + 1 \times -2}{3 + 1}\Big) \\[1em] = \Big(\dfrac{27 + 5}{4}, \dfrac{18 - 2}{4}\Big) \\[1em] = \Big(\dfrac{32}{4}, \dfrac{16}{4}\Big) \\[1em] = (8, 4).$

Hence, Option 3 is the correct option.

The coordinates of the point on x-axis which divides the line segment joining the points (2, 3) and (5, -6) in the ratio 1 : 2 are :

1. (2, 0)

2. (-2, 0)

3. (3, 0)

4. (-3, 0)

Answer

Let point P be (x, y).

Given,

m₁ : m₂ = 1 : 2

By section-formula,

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

Substituting values we get :

$\Rightarrow (x, y) = \Big(\dfrac{1 \times 5 + 2 \times 2}{1 + 2}, \dfrac{1 \times (-6) + 2 \times 3}{1 + 2}\Big) \\[1em] = \Big(\dfrac{5 + 4}{3}, \dfrac{-6 + 6}{3}\Big) \\[1em] = \Big(\dfrac{9}{3}, \dfrac{0}{3}\Big) \\[1em] = (3, 0).$

Hence, Option 3 is the correct option.

The point which divides the line segment joining the points A(3, -2) and B(6, 7) internally in the ratio 3 : 2 lies in which of the following quadrants?

1. I

2. II

3. III

4. IV

Answer

Let point P be (x, y).

Given,

m₁ : m₂ = 3 : 2

By section-formula,

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

Substituting values we get :

$\Rightarrow (x, y) = \Big(\dfrac{3 \times 6 + 2 \times 3}{3 + 2}, \dfrac{3 \times 7 + 2 \times (-2)}{3 + 2}\Big) \\[1em] = \Big(\dfrac{18 + 6}{5}, \dfrac{21 - 4}{5}\Big) \\[1em] = \Big(\dfrac{24}{5}, \dfrac{17}{5}\Big).$

Here, both x and y are positive.

Therefore, the point lies in the 1st Quadrant.

Hence, Option 1 is the correct option.

If the point R(k, 4) divides the line segment joining the points P(2, 6) and Q(5, 1) in the ratio 2 : 3, then the value of k is:

1. -5

2. $\Big(\dfrac{-16}{5}\Big)$

3. 5

4. $\Big(\dfrac{16}{5}\Big)$

Answer

Given,

R = (k, 4)

m₁ : m₂ = 2 : 3

R(k, 4) divides the line segment joining the points P(2, 6) and Q(5, 1) in the ratio 2 : 3.

By section-formula,

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

Substituting values we get :

$\Rightarrow (k, 4) = \Big(\dfrac{2 \times 5 + 3 \times 2}{2 + 3}, \dfrac{2 \times 1 + 3 \times 6}{2 + 3}\Big) \\[1em] \Rightarrow (k, 4) = \Big(\dfrac{10 + 6}{5}, \dfrac{2 + 18}{5}\Big) \\[1em] \Rightarrow (k, 4) = \Big(\dfrac{16}{5}, \dfrac{20}{5}\Big) \\[1em] \Rightarrow (k, 4) = \Big(\dfrac{16}{5}, 4\Big).$

Thus, k = $\dfrac{16}{5}$.

Hence, Option 4 is the correct option.

Points A(x, y), B(3, -2) and C(4, -5) are collinear. The value of y in terms of x is ∶

1. 3x - 11

2. 11 - 3x

3. 3x - 7

4. 7 - 3x

Answer

Since, points A, B and C are collinear.

∴ Slope of AB = Slope of BC.

$\Rightarrow \dfrac{-2 - y}{3 - x} = \dfrac{-5 - (-2)}{4 - 3} \\[1em] \Rightarrow \dfrac{-2 - y}{3 - x} = \dfrac{-5 + 2}{1} \\[1em] \Rightarrow \dfrac{-2 - y}{3 - x} = -3 \\[1em] \Rightarrow -2 - y = -3(3 - x) \\[1em] \Rightarrow -2 - y = -9 + 3x \\[1em] \Rightarrow y = -2 + 9 - 3x \\[1em] \Rightarrow y = 7 - 3x.$

Hence, Option 4 is the correct option.

If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3 : 1, then the value of y is :

1. 1

2. 2

3. 3

4. 4

Answer

Let point P be (6, 2).

Given,

m₁ : m₂ = 3 : 1

P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3 : 1.

By section-formula,

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

We use the y–coordinate to find y.

$\Rightarrow 2 = \dfrac{3 \times y + 1 \times 5}{3 + 1} \\[1em] \Rightarrow 2 = \dfrac{3y + 5}{4} \\[1em] \Rightarrow 8 = 3y + 5 \\[1em] \Rightarrow 3y = 3 \\[1em] \Rightarrow y = 1.$

Hence, Option 1 is the correct option.

The ratio in which the point P(1, 2) divides the join of the points A(-2, 1) and B(7, 4) is:

1. 1 : 2

2. 2 : 1

3. 3 : 2

4. 2 : 3

Answer

Let the ratio in which P divides AB be k : 1.

By section-formula,

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

Using the y–coordinate to find ratio.

$\Rightarrow 2 = \Big(\dfrac{k(4) + 1(1)}{k + 1}\Big) \\[1em] \Rightarrow 2 = \Big(\dfrac{4k + 1}{k + 1}\Big) \\[1em] \Rightarrow 2(k + 1) = 4k + 1 \\[1em] \Rightarrow 2k + 2 = 4k + 1 \\[1em] \Rightarrow 2 - 1 = 4k - 2k \\[1em] \Rightarrow 2k = 1 \\[1em] \Rightarrow k = \dfrac{1}{2}\\[1em] \Rightarrow k : 1 = \dfrac{1}{2} : 1 = 1 : 2.$

Hence, Option 1 is the correct option.

The line segment joining A(-7, 2) and B(3, -8) is divided by the x-axis in the ratio:

1. 1 : 4

2. 3 : 7

3. 4 : 1

4. 7 : 3

Answer

Given,

AB is divided by the x-axis, thus y-coordinate = 0 at point of division.

A(-7, 2) and B(3, -8)

Let ratio be m : n.

By section-formula,

y = $\dfrac{my_2 + ny_1}{m + n}$

Substituting values we get :

⇒ 0 = $\dfrac{m \times (-8) + n \times (2)}{m + n}$

⇒ 0 = -8m + 2n

⇒ 8m = 2n

⇒ $\dfrac{m}{n} = \dfrac{2}{8}$

⇒ $\dfrac{m}{n} = \dfrac{1}{4}$

⇒ m : n = 1 : 4.

Hence, option 1 is the correct option.

In what ratio is the line segment joining the points P(-4, 2) and Q(8, 3) divided by y-axis?

1. 1 : 3

2. 3 : 1

3. 1 : 2

4. 2 : 1

Answer

Let the point where y-axis divides the line segment be R(0, y).

Let the ratio be m₁ : m₂.

Using section-formula,

$\Rightarrow x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 0 = \dfrac{m_1 \times 8 + m_2 \times (-4)}{m_1 + m_2} \\[1em] \Rightarrow 0 = 8m_1 - 4m_2 \\[1em] \Rightarrow 8m_1 = 4m_2 \\[1em] \Rightarrow \dfrac{m_1}{m_2} = \dfrac{4}{8} = \dfrac{1}{2}.$

Thus, the required ratio is 1 : 2.

Hence, Option 3 is the correct option.

Point P divides the line segment joining R(-1, 3) and S(9, 8) in the ratio k : 1. If P lies on the line x - y + 2 = 0, then the value of k is:

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

2. $\dfrac{1}{3}$

3. $\dfrac{1}{4}$

4. $\dfrac{2}{3}$

Answer

Let point P be (x, y).

Given,

m₁ : m₂ = k : 1

By section-formula,

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

Substituting values we get :

$\Rightarrow (x, y) = \Big(\dfrac{k \times 9 + 1 \times (-1)}{k + 1}, \dfrac{k \times 8 + 1 \times 3}{k + 1}\Big) \\[1em] = \Big(\dfrac{9k - 1}{k + 1}, \dfrac{8k + 3}{k + 1}\Big).$

Since P lies on the line x - y + 2 = 0, substituting values of x and y:

$\Rightarrow \dfrac{9k - 1}{k + 1} - \dfrac{8k + 3}{k + 1} + 2 = 0 \\[1em] \Rightarrow \dfrac{9k - 1 - (8k + 3)}{k + 1} + 2 = 0 \\[1em] \Rightarrow \dfrac{9k - 1 - 8k - 3}{k + 1} + 2 = 0 \\[1em] \Rightarrow \dfrac{k - 4}{k + 1} + 2 = 0 \\[1em] \Rightarrow \dfrac{k - 4 + 2(k + 1)}{k + 1} = 0 \\[1em] \Rightarrow \dfrac{k - 4 + 2k + 2}{k + 1} = 0 \\[1em] \Rightarrow \dfrac{3k - 2}{k + 1} = 0 \\[1em] \Rightarrow 3k - 2 = 0 \\[1em] \Rightarrow 3k = 2 \\[1em] \Rightarrow k = \dfrac{2}{3}.$

Hence, Option 4 is the correct option.

In the adjoining figure, P(5, -3) and Q(3, y) are the points of trisection of the line segment joining A(7, -2) and B(1, -5). Then, y equals :

1. -4

2. $\Big(\dfrac{-5}{2}\Big)$

3. 2

4. 4

Answer

Since P and Q trisect the line segment AB, the point Q(3, y) divides A(7, -2) and B(1, -5) in the ratio 2 : 1.

Let point Q be (3, y).

Given,

m₁ : m₂ = 2 : 1

By section-formula,

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

Substituting values we get :

$\Rightarrow (3, y) = \Big(\dfrac{2 \times 1 + 1 \times 7}{2 + 1}, \dfrac{2 \times (-5) + 1 \times (-2)}{2 + 1}\Big) \\[1em] \Rightarrow (3, y) = \Big(\dfrac{2 + 7}{3}, \dfrac{-10 - 2}{3}\Big) \\[1em] \Rightarrow (3, y) = \Big(\dfrac{9}{3}, \dfrac{-12}{3}\Big) \\[1em] \Rightarrow (3, y) = (3, -4).$

Thus, y = -4.

Hence, Option 1 is the correct option.

If the point P(6, -3) lies on the line segment joining points A(4, 2) and B(8, 4), then:

1. AP = $\dfrac{3}{4}$ AB

2. AP = $\dfrac{1}{4}$ AB

3. PB = $\dfrac{1}{3}$ AB

4. AP = $\dfrac{1}{2}$ AB

Answer

Let the ratio in which P divides AB be k : 1.

By section-formula,

x = $\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get :

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

This means that P is the midpoint of AB.

∴ AP = $\dfrac{1}{2}$ AB.

Hence, Option 4 is the correct option.

The mid-point of the line segment joining the points (-3, 2) and (7, 6) is:

1. (-2, -4)

2. (-2, 4)

3. (2, 4)

4. (4, 2)

Answer

Let the mid-point be M(x, y).

By mid-point formula,

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

Substituting values we get :

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

Hence, Option 3 is the correct option.

If A(4, 2), B(6, 5) and C(1, 4) be the vertices of ΔABC and AD is a median, then the coordinates of D are:

1. $\Big(\dfrac{5}{2}, 3\Big)$

2. $\Big(5, \dfrac{7}{2}\Big)$

3. $\Big(\dfrac{7}{2}, \dfrac{9}{2}\Big)$

4. none of these

Answer

Since AD is a median, D is the mid-point of BC.

Let point D be (x, y).

By mid-point formula,

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

$\Rightarrow (x, y) = \Big(\dfrac{6 + 1}{2}, \dfrac{5 + 4}{2}\Big) \\[1em] \Rightarrow \Big(\dfrac{7}{2}, \dfrac{9}{2}\Big).$

Hence, Option 3 is the correct option.

If (3, -6) is the mid-point of the line segment joining (0, 0) and (x, y), then the point (x, y) is:

1. (-3, 6)

2. (6, -6)

3. (6, -12)

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

Answer

Given, (3, -6) is the mid-point of the line segment joining (0, 0) and (x, y).

$\therefore (3, -6) = \Big(\dfrac{0 + x}{2}, \dfrac{0 + y}{2}\Big) \\[1em] \Rightarrow (3, -6) = \Big(\dfrac{x}{2}, \dfrac{y}{2}\Big) \\[1em] \Rightarrow 3 = \dfrac{x}{2} \text{ and } -6 = \dfrac{y}{2} \\[1em] \Rightarrow x = 6 \text{ and } y = -12$

(x, y) = (6, -12).

Hence, Option 3 is the correct option.

If the line segment joining the points P and Q(3, -4) is bisected at the origin, then the coordinates of P are:

1. (3, -2)

2. (3, -4)

3. (-3, -4)

4. (-3, 4)

Answer

Let the coordinates of P be (x, y).

Given, the origin (0, 0) is the mid-point of PQ.

By mid-point formula,

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

Substituting values we get :

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

Hence, Option 4 is the correct option.

A(-3, b) and B(1, b + 4) are two points. If the coordinates of the mid-point of AB are (-1, 1), then the value of b is :

1. -1

2. 0

3. 1

4. 2

Answer

Given,

Mid-point of AB = (-1, 1).

By mid-point formula,

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

Substituting values we get :

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

Comparing the y-coordinates, we get :

⇒ 1 = b + 2

⇒ b = -1.

Hence, Option 1 is the correct option.

If the point R(5, 7) is the mid-point of the line segment joining the points P(3, y) and Q(x, 9), then (x + y) equals:

1. 7

2. 9

3. 12

4. 14

Answer

Given,

R(5, 7) is the mid-point of the line segment joining the points P(3, y) and Q(x, 9).

By mid-point formula,

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

Substituting values we get :

$\Rightarrow (5, 7) = \Big(\dfrac{3 + x}{2}, \dfrac{y + 9}{2}\Big) \\[1em] \Rightarrow 5 = \dfrac{3 + x}{2} \text{ and } 7 = \dfrac{y + 9}{2} \\[1em] \Rightarrow 10 = 3 + x \text{ and } 14 = y + 9 \\[1em] \Rightarrow x = 7 \text{ and } y = 5.$

x + y = 7 + 5 = 12.

Hence, Option 3 is the correct option.

The mid-point of the line segment joining (4p, 5) and (2, 3q) is (5, 5p - 1). The values of p and q are respectively:

1. 2, $\dfrac{8}{3}$

2. -2, $\dfrac{8}{3}$

3. 2, $\dfrac{13}{3}$

4. -2, $\dfrac{13}{3}$

Answer

Given,

Mid-point of the line segment joining (4p, 5) and (2, 3q) is (5, 5p - 1).

By mid-point formula,

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

Substituting values we get :

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

Comparing the x coordinates, we get :

⇒ 2p + 1 = 5

⇒ 2p = 5 - 1

⇒ 2p = 4

⇒ p = $\dfrac{4}{2}$

⇒ p = 2.

Comparing y-coordinates we get :

$\Rightarrow 5p - 1 = \dfrac{5 + 3q}{2} \\[1em] \Rightarrow 5 \times 2 - 1 = \dfrac{5 + 3q}{2} \\[1em] \Rightarrow 9 = \dfrac{5 + 3q}{2} \\[1em] \Rightarrow 18 = 5 + 3q \\[1em] \Rightarrow 3q = 18 - 5 \\[1em] \Rightarrow 3q = 13 \\[1em] \Rightarrow q = \dfrac{13}{3}.$

p = 2 and q = $\dfrac{13}{3}$.

Hence, Option 3 is the correct option.

A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid-point of PQ, then the coordinates of P and Q are respectively:

1. (0, 10) and (-4, 0)

2. (0, -5) and (2, 0)

3. (0, 4) and (-10, 0)

4. (0, -10) and (4, 0)

Answer

Let P(0, a) be the point on y-axis and Q(b, 0) be the point on x-axis.

Given, (2, -5) is the mid-point of PQ.

By mid-point formula,

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

Substituting values we get :

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

P(0, -10) and Q(4, 0)

Hence, Option 4 is the correct option.

If a point $R\Big(\dfrac{23}{5}, \dfrac{33}{5}\Big)$ divides the line segment PQ joining the points P(3, 5) and Q(x, y) in the ratio 2 : 3 internally, then the values of x and y respectively are :

1. 4, 7

2. 5, 9

3. 7, 8

4. 7, 9

Answer

Given,

Point R = $\Big(\dfrac{23}{5}, \dfrac{33}{5}\Big)$ and P(3, 5), Q(x, y).

Given,

m₁ : m₂ = 2 : 3

By section-formula,

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

Substituting values we get :

$\Rightarrow \Big(\dfrac{23}{5}, \dfrac{33}{5}\Big) = \Big(\dfrac{2x + 3 \times 3}{2 + 3}, \dfrac{2y + 3 \times 5}{2 + 3}\Big) \\[1em] \Rightarrow \dfrac{23}{5} = \dfrac{2x + 9}{5}, \dfrac{33}{5} = \dfrac{2y + 15}{5} \\[1em] \Rightarrow 23 = 2x + 9, 33 = 2y + 15 \\[1em] \Rightarrow 23 - 9 = 2x, 33 - 15 = 2y \\[1em] \Rightarrow 14 = 2x, 18 = 2y \\[1em] \Rightarrow x = \dfrac{14}{2}, y = \dfrac{18}{2} \\[1em] \Rightarrow x = 7, y = 9.$

Therefore, the values of x and y are 7 and 9 respectively.

Hence, Option 4 is the correct option.

The coordinates of the vertices of ΔABC are respectively (-4, -2), (6, 2) and (4, 6). The centroid G of ΔABC is:

1. (2, 2)

2. (2, 3)

3. (3, 3)

4. (0, -1)

Answer

By centroid formula,

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

Substituting values we get :

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

Hence, Option 1 is the correct option.

Two vertices of a ΔABC are A(-1, 4) and B(5, 2) and its centroid is (0, -3). The coordinates of C are :

1. (4, 3)

2. (4, 15)

3. (-4, -15)

4. (-15, -4)

Answer

Let the coordinates of C be (x, y).

By centroid formula,

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

Given,

G(0, -3) is the centroid of the triangle.

Substituting values we get :

$\Rightarrow (0, -3) = \Big(\dfrac{-1 + 5 + x}{3}, \dfrac{4 + 2 + y}{3}\Big) \\[1em] \Rightarrow (0, -3) = \Big(\dfrac{4 + x}{3}, \dfrac{6 + y}{3}\Big) \\[1em] \Rightarrow 0 = \dfrac{4 + x}{3} \text{ and } -3 = \dfrac{6 + y}{3} \\[1em] \Rightarrow 4 + x = 0 \text{ and } 6 + y = -9 \\[1em] \Rightarrow x = -4 \text{ and } y = -9 - 6 \\[1em] \Rightarrow x = -4 \text{ and } y = -15.$

C = (x, y) = (-4, -15).

Hence, Option 3 is the correct option.

In the adjoining diagram, G is the centroid of △ ABC. A(3, -3), B(2, -6), C(x, y) and G(5, -5). The coordinates of point D are :

1. (2, -6)

2. (3, -6)

3. (6, -6)

4. (10, -6)

Answer

By formula,

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

$\therefore (5, -5) = \Big(\dfrac{3 + 2 + x}{3}, \dfrac{(-3) + (-6) + y}{3}\Big) \\[1em] \Rightarrow (5, -5) = \Big(\dfrac{x + 5}{3}, \dfrac{y - 9}{3}\Big) \\[1em] \Rightarrow \dfrac{x + 5}{3} = 5 \text{ and } \dfrac{y - 9}{3} = -5 \\[1em] \Rightarrow x + 5 = 15 \text{ and } y - 9 = -15 \\[1em] \Rightarrow x = 15 - 5 = 10 \text{ and } y = -15 + 9 = -6.$

C(x, y) = (10, -6).

Since, centroid is the point of intersection of all the three medians of a triangle.

∴ AD is the median.

∴ D is mid-point of BC.

$D = \Big(\dfrac{2 + 10}{2}, \dfrac{(-6) + (-6)}{2}\Big) \\[1em] = \Big(\dfrac{12}{2}, \dfrac{-12}{2}\Big) \\[1em] = (6, -6).$

Hence, Option 3 is the correct option.

The points A, B and C divide the line segment joining the points P(-3, 8) and Q(9, -4) into four equal parts. If A is nearest to P, then the coordinates of A are:

1. (-3, 5)

2. (0, 5)

3. (3, 5)

4. (6, -1)

Answer

Given,

The points dividing PQ into four equal parts be A, B and C such that A is nearest to P.

Since the line segment PQ is divided into four equal parts, the point A divides PQ in the ratio 1 : 3.

By section-formula,

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

Substituting values we get :

$\Rightarrow (x, y) = \Big(\dfrac{1 \times 9 + 3 \times (-3)}{1 + 3}, \dfrac{1 \times (-4) + 3 \times 8}{1 + 3}\Big) \\[1em] \Rightarrow \Big(\dfrac{9 - 9}{4}, \dfrac{-4 + 24}{4}\Big) \\[1em] \Rightarrow \Big(\dfrac{0}{4}, \dfrac{20}{4}\Big) \\[1em] \Rightarrow (0, 5).$

Hence, Option 2 is the correct option.

The line 2x + y - 4 = 0 divides the line segment joining A(2, -2) and B(3, 7) in the ratio:

1. 2 : 3

2. 2 : 5

3. 2 : 7

4. 2 : 9

Answer

Let the required point be P(x, y) which divides A(2, -2) and B(3, 7) in the ratio k : 1.

By section-formula,

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

Substituting values, we get :

$\Rightarrow (x, y) = \Big(\dfrac{k \times 3 + 1 \times 2}{k + 1}, \dfrac{k \times 7 + 1 \times (-2)}{k + 1}\Big) \\[1em] \Rightarrow (x, y) = \Big(\dfrac{3k + 2}{k + 1}, \dfrac{7k - 2}{k + 1}\Big)$

Since P lies on the line 2x + y - 4 = 0, substituting the values of x and y:

$\Rightarrow 2\Big(\dfrac{3k + 2}{k + 1}\Big) + \Big(\dfrac{7k - 2}{k + 1}\Big) - 4 = 0 \\[1em] \Rightarrow \dfrac{6k + 4 + 7k - 2}{k + 1} - 4 = 0 \\[1em] \Rightarrow \dfrac{13k + 2}{k + 1} = 4 \\[1em] \Rightarrow 13k + 2 = 4(k + 1) \\[1em] \Rightarrow 13k + 2 = 4k + 4 \\[1em] \Rightarrow 9k = 2 \\[1em] \Rightarrow k = \dfrac{2}{9} \\[1em] \Rightarrow k : 1 = \dfrac{2}{9} : 1 = 2 : 9.$

Hence, Option 4 is the correct option.

The centre of the circle having end points of its one diameter as (-4, 2) and (4, -3) is:

1. (0, -1)

2. (2, -1)

3. $\Big(0, -\dfrac{1}{2}\Big)$

4. $\Big(4, -\dfrac{5}{2}\Big)$

Answer

Let the end points of the diameter be A(-4, 2) and B(4, -3). The centre of the circle is the mid-point of the diameter AB.

By mid-point formula,

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

Substituting values we get :

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

Hence, Option 3 is the correct option.

A circle has its centre at (4, 4). If one end of a diameter is (4, 0), then the coordinates of the other end are:

1. (0, 4)

2. (4, 8)

3. (4, -8)

4. (-4, -8)

Answer

Let one end of the diameter be A(4, 0) and the other end be B(x, y). Given that the centre of the circle is (4, 4), which is the mid-point of AB.

By mid-point formula,

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

Substituting values we get :

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

The coordinates of the other end of the diameter are (4, 8).

Hence, Option 2 is the correct option.

The vertices of a parallelogram in order are A(1, 2), B(4, y), C(x, 6) and D(3, 5). Then (x, y) is:

1. (6, 3)

2. (3, 6)

3. (5, 6)

4. (1, 4)

Answer

In a parallelogram, the diagonals bisect each other. Therefore, the mid-point of AC = mid-point of BD.

By mid-point formula,

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

Substituting values, we get :

For diagonal AC :

$\text{Mid-point of AC} = \Big(\dfrac{1 + x}{2}, \dfrac{2 + 6}{2}\Big) = \Big(\dfrac{1 + x}{2}, 4\Big)$

For diagonal BD:

$\text{Mid-point of BD} = \Big(\dfrac{4 + 3}{2}, \dfrac{y + 5}{2}\Big) = \Big(\dfrac{7}{2}, \dfrac{y + 5}{2}\Big)$

Since both mid-points are equal, we equate their coordinates:

$\Rightarrow \Big(\dfrac{1 + x}{2}, 4\Big) = \Big(\dfrac{7}{2}, \dfrac{y + 5}{2}\Big) \\[1em] \Rightarrow \dfrac{1 + x}{2} = \dfrac{7}{2}, 4 = \dfrac{y + 5}{2}\\[1em] \Rightarrow 1 + x = 7, y + 5 = 8 \\[1em] \Rightarrow x = 7 - 1, y = 8 - 5 \\[1em] \Rightarrow x = 6, y = 3.$

(x, y) = (6, 3).

Hence, Option 1 is the correct option.

The fourth vertex D of a parallelogram ABCD whose three 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

In a parallelogram, the diagonals bisect each other. Therefore, the mid-point of AC = mid-point of BD.

By mid-point formula,

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

Substituting values, we get :

For diagonal AC:

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

Let point D be (x, y).

For diagonal BD:

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

D = (x, y) = (0, -1).

Hence, Option 2 is the correct option.

A(1, 4), B(4, 1) and C(x, 4) are the vertices of ΔABC. If the centroid of the triangle is G(4, 3), then x is equal to:

1. 2

2. 1

3. 7

4. 4

Answer

Given,

The vertices of the triangle be A(1, 4), B(4, 1) and C(x, 4). The centroid G of a triangle is given by the formula:

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

Substituting the given values we get :

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

Hence, Option 3 is the correct option.