This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Geometry – Section Formula”.

1. What will be the coordinates of the point which divides the line segment joining the points A(-2, 2) and B(-1, 5) in the ratio 2:5?

a) (\(\frac {-4}{3}, \frac {-20}{9}\))

b) (\(\frac {-4}{3}, \frac {20}{9}\))

c) (\(\frac {4}{3}, \frac {20}{9}\))

d) (\(\frac {4}{3}, \frac {-20}{9}\))

View Answer

Explanation: Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(-2, 2) and B(-1, 5) and the ratio is 2:5

∴ x = \(\frac {2(-1)+5(-2)}{2+7} = \frac {-2-10}{9} = \frac {-12}{9} = \frac {-4}{3}\)

y = \(\frac {2(5)+5(2)}{2+7} = \frac {10+10}{9} = \frac {20}{9} = \frac {20}{9}\)

Hence, the point is (\(\frac {-4}{3}, \frac {20}{9}\)).

2. What will be the coordinates of the midpoint of the line segment joining the points (-5, 10) and(15, 2)?

a) (-5, -6)

b) (-5, 6)

c) (5, 6)

d) (5, -6)

View Answer

Explanation: Midpoint lies in the center of the line segment

Hence, it divides the line in the ratio 1:1

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(-5, 10) and B(15, 2) and the ratio is 1:1

∴ x = \(\frac {1(-5)+1(15)}{2} = \frac {-5+15}{2} = \frac {10}{2}\) = 5

y = \(\frac {1(2)+1(10)}{2} = \frac {2+10}{2} = \frac {12}{2}\) = 6

Hence, the point is (5,6).

3. In what ratio does the point (\(\frac {-19}{3}, \frac {7}{3}\)) divide the line segment joining A(3, 7) and B(-11, 0)?

a) 1:2 (externally)

b) 1:2 (internally)

c) 2:1 (externally)

d) 2:1 (internally)

View Answer

Explanation: Let the ratio in which the point (\(\frac {-19}{3}, \frac {7}{3}\)) divides the line segment joining the points A(3, 7) and B(-11, 0) be k:1

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(3, 7) and B(-11, 0) and the ratio is k:1

∴ x = \(\frac {k(-11)+1(3)}{k+1} = \frac {-11k+3}{k+1}\)

y = \(\frac {k(0)+1(7)}{k+1} = \frac {7}{k+1}\)

Since, the point is (\(\frac {-19}{3}, \frac {7}{3}\)).

∴ \(\frac {-19}{3} = \frac {-11k+3}{k+1}\)

-19(k + 1) = 3(-11k + 3)

-19k – 19 = -33k + 9

-19k + 33k = 19 + 9

14k = 28

k = \(\frac {28}{14}\) = 2

The ratio is 2:1.

4. What will be the value of y, if the ratio in which the point (\(\frac {3}{4}\), y) divides the line segment joining the points A(-1, 4) and B(6, 5)is 1:3?

a) y = \(\frac {9}{2}\)

b) y = \(\frac {5}{2}\)

c) y = \(\frac {9}{4}\)

d) y = \(\frac {5}{2}\)

View Answer

Explanation: Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are (-1, 4)and B(6, 5)in the ratio 1:3

∴ x = \(\frac {1(6)+3(-1)}{1+3} = \frac {6-3}{4} = \frac {3}{4}\)

y = \(\frac {1(6)+3(4)}{1+3} = \frac {6+12}{4} = \frac {18}{4}\)

Therefore y = \(\frac {9}{2}\)

5. What will be ratio in which the line 3x + y – 11 = 0 divides the line segment joining the points (0, -1) and (-3, -4)?

a) 1:2 (internally)

b) 1:2 (externally)

c) 2:1 (externally)

d) 2:1 (internally)

View Answer

Explanation: Let the ratio in which the line 3x + y – 11 = 0 divides the line segment joining the points (0, -1) and (-3, -4) be k:1.

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(0, -1) and B(-3, -4) and the ratio is k:1.

∴ x = \(\frac {k(-3)+1(0)}{k+1} = \frac {-3k}{k+1}\)

y = \(\frac {k(-4)+1(-1)}{k+1} = \frac {-4k-1}{k+1}\)

Since, the point \((\frac {-3k}{k+1}, \frac {-4k-1}{k+1} )\) lies on the line 3x+y-11 = 0.

3 \((\frac {-3k}{k+1} + \frac {-4k-1}{k+1} )\)-11 = 0

3(-3k) + (-4k – 1) – 11(k + 1) = 0

-9k – 4k – 1 – 11k – 11 = 0

-24k – 12 = 0

-24k = 12

k = \(\frac {12}{-24} = \frac {-1}{2}\)

The ratio is 1:2 (externally).

6. In what ratio is the line segment joining the points A(-5, 2) and B(3, 9) divided by the x-axis?

a) 2:5 (internally)

b) 2:5 (externally)

c) 2:9 (externally)

d) 2:9 (internally)

View Answer

Explanation: Let the ratio in which the x-axis divides the line segment joining the points A(-5, 2) and B(3, 9) be k:1

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(-5, 2) and B(3, 9) and the ratio is k:1

∴ x = \(\frac {k(3)+1(-5)}{k+1} = \frac {3k-5}{k+1}\)

y = \(\frac {k(9)+1(2)}{k+1} = \frac {9k+2}{k+1}\)

Since, the point is on x-axis.

Hence, the y-coordinate will be zero.

∴ 0 = \(\frac {9k+2}{k+1}\)

0 = 9k+2

k = \(\frac {-2}{9}\)

The ratio in which the y-axis cuts the line segment joining the points A(-5, 2) and B(3, 9) will be 2:9 (externally).

7. In what ratio is the line segment joining the points A(2, 4) and B(6, 5) divided by the y-axis?

a) 2:1 (internally)

b) 2:1 (externally)

c) 3:1 (internally)

d) 3:1 (externally)

View Answer

Explanation: Let the ratio in which the y-axis divides the line segment joining the points A(2, 4) and B(6, 5) be k:1

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(2, 4) and B(6, 5) and the ratio is k:1

∴ x = \(\frac {k(6)+1(2)}{k+1} = \frac {6k+2}{k+1}\)

y = \(\frac {k(5)+1(4)}{k+1} = \frac {5k+4}{k+1}\)

Since, the point is on y-axis.

Hence, the x-coordinate will be zero.

∴ 0 = \(\frac {6k+2}{k+1}\)

0 = 6k + 2

k = \(\frac {-6}{2}\) = -3

The ratio in which the y-axis cuts the line segment joining the points A(2, 4) and B(6, 5) will be 3:1 (externally).

8. What will be the coordinates of B, if the point C\((\frac {29}{7}, \frac {46}{7} )\), divides the line segment joining A (5, 8) and B (a, b) in the ratio 2:5?

a) a = 2, b = 3

b) a = -2, b = 3

c) a = 2, b = -3

d) a = -2, b = -3

View Answer

Explanation: Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(5, 8)and B(a, b)in the ratio 2:5

∴ x = \(\frac {2(a)+5(5)}{2+5} = \frac {2a+25}{7}\)

y = \(\frac {2(b)+5(8)}{2+5} = \frac {2b+40}{7}\)

But the coordinates of C are \((\frac {29}{7}, \frac {46}{7} )\)

Therefore, \(\frac {2a+25}{7} = \frac {29}{7}\)

a = 2

\(\frac {2b+40}{7} = \frac {46}{7}\)

b = 3

9. What will be the length of the median through the vertex A, if the coordinates of the vertices of ∆ABC are A(2, 5), B(5, 0), C(-2, 5)?

a) \(\sqrt {\frac {113}{3}}\) units

b) \(\sqrt {\frac {13}{2}}\) units

c) \(\sqrt {\frac {113}{2}}\) units

d) \(\sqrt {\frac {13}{2}}\) units

View Answer

Explanation:

The median through A will bisect the line BC.

Hence, D is the midpoint of BC

Coordinates of D = \((\frac {x_1+x_2}{2}, \frac {y_1+y_2}{2} ) = ( \frac {5-2}{2}, \frac {0-5}{2} ) =( \frac {3}{2}, \frac {-5}{2} )\)

Distance between A and D = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(2-\frac {3}{2})^2+ (5+\frac {5}{2})^2} \)

= \( \sqrt {(\frac {1}{2})^2+ (\frac {15}{2})^2} \)

= \( \sqrt {\frac {1}{4}+ \frac {225}{4}} \)

= \( \sqrt {\frac {113}{2}} \) units

10. What will be the coordinates of the fourth vertex S, if P(-1, -1), Q(2, 0), R(2, 3) are the three vertices of a parallelogram?

a) (-5, -12)

b) (5, -12)

c) (5, 12)

d) (-5, 12)

View Answer

Explanation:

PQRS is a parallelogram. The opposite side of the parallelogram is equal and parallelogram. Also, the diagonals of the parallelogram bisect each other.

∴ O is the mid-point SQ and PR.

Midpoint of PR

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are P(-1, -1) and R(2, 3) and the ratio is 1:1

∴ x = \(\frac {1(-1)+1(2)}{2} = \frac {-1+2}{2} = \frac {1}{2}\)

y = \(\frac {1(3)+1(-1)}{2} = \frac {3-1}{2} = \frac {2}{2}\) = 1

Hence, the coordinates of O is (5, 6)

Midpoint of QS

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are Q(2, 0) and S(a, b) and the ratio is 1:1

∴ x = \(\frac {1(a)+1(2)}{2} = \frac {a+2}{2}\)

y = \(\frac {1(b)+1(0)}{2} = \frac {b}{2}\)

The coordinates of O is (5, 6)

Therefore, \(\frac {a+2}{2}\) = 5

a = 8

\(\frac {b}{2}\) = 6, b = 12

The coordinates of S are (5, 12).

11. What will be the value of a and b, if (-5, a), (-3, -3), (-b, 0) and (-3, 3) are the vertices of the parallelogram?

a) a = 0, b = -1

b) a = -1, b = 1

c) a = 1, b = 1

d) a = 0, b = 1

View Answer

Explanation:

PQRS is a parallelogram. The opposite side of the parallelogram is equal and parallelogram. Also, the diagonals of the parallelogram bisect each other.

∴ O is the mid-point SQ and PR.

**Midpoint of PR**

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are P(-5, a) and R(-b, 0) and the ratio is 1:1

∴ x = \(\frac {1(-b)+1(-5)}{2} = \frac {-b-5}{2}\)

y = \(\frac {1(0)+1(a)}{2} = \frac {a}{2}\)

**Midpoint of QS**

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are Q(-3, -3) and S(-3, 3) and the ratio is 1:1

∴ x = \(\frac {1(-3)+1(-3)}{2} = \frac {-6}{2}\) = -3

y = \(\frac {1(3)+1(-3)}{2} = \frac {0}{2}\) = 0

Therefore, \(\frac {-b-5}{2}\) = -3

b = 1

\(\frac {a}{2}\) = 0

a = 0

12. What will be the centroid of the ∆ABC whose vertices are A(-2, 4), B(0, 0) and C(4, 2)?

a) (\(\frac {2}{3}\), 2)

b) (\(\frac {2}{3}\), 1)

c) (\(\frac {2}{5}\), 2)

d) (\(\frac {1}{3}\), 2)

View Answer

Explanation: We know, x

_{centroid}= \(\frac {x_1+x_2+x_3}{3}\) and y

_{centroid}= \(\frac {y_1+y_2+y_3}{3}\)

x

_{centroid}= \(\frac {-2+0+4}{3} = \frac {2}{3}\)

y

_{centroid}= \(\frac {4+0+2}{3}\) = 2

The coordinates of the centroid are (\(\frac {2}{3}\), 2).

13. The coordinates of one end of the diameter AB of a circle are A (-2, -3) and the coordinates of diameter are (-2, 0). What will be the coordinates of B?

a) (2, -3)

b) (-2, 3)

c) (2, 3)

d) (-2, -3)

View Answer

Explanation: We know that the diameter is twice the radius.

Hence, the center is the midpoint of the diameter.

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(-2, -3) and center is (-2, 0) and the ratio is 1:1

Let the coordinates of other side of the radius be (x, y).

∴ -2 = \(\frac {1(-2)+1(x)}{2} = \frac {-2+x}{2}\)

-4 = -2 + x

-4 + 2 = x

x = -2

0 = \(\frac {1(-3)+1(y)}{2} = \frac {-3+y}{2}\)

0 = -3 + y

y = 3

Hence, the point is (-2, 3).

14. The coordinates of the ends of the diameter AB of a circle are A (-4, 7) and B(4, 7). What will be the coordinates of the center of the circle?

a) (0, -8)

b) (0, 8)

c) (0, 7)

d) (0, -7)

View Answer

Explanation: We know that the diameter is twice the radius.

Hence, the center is the midpoint of the diameter.

Using, section formula x = \(\frac {mx_2+nx_1}{m+n}\) and y = \(\frac {my_2+ny_1}{m+n}\)

The points are A(-4, 7) and B(4, 7) and the ratio is 1:1

∴ x = \(\frac {1(-4)+1(4)}{2} = \frac {0}{2}\) = 0

y = \(\frac {1(7)+1(7)}{2} = \frac {7+7}{2} = \frac {14}{2}\) = 7

Hence, the point is (0, 7).

15. The two vertices of ∆ABC are given by A(-3, 0) and B(-8, 5) and its centroid is (-2, 1).What will be the coordinates of the third vertex C?

a) (-5, -2)

b) (5, 2)

c) (-5, 2)

d) (5, -2)

View Answer

Explanation: The two vertices of triangle are A (-3, 0) and B (-8, 5). Its centroid is (-2, 1).

We know, x

_{centroid}= \(\frac {x_1+x_2+x_3}{3}\) and y

_{centroid}= \(\frac {y_1+y_2+y_3}{3}\)

Now, x

_{centroid}= \(\frac {-3-8+x_3}{3}\)

x

_{centroid}= -2

-2 = \(\frac {-3 – 8 + x_3}{3}\)

-6 = -3 – 8 + x

_{3}

5 = x

_{3}

Now, y

_{centroid}= \(\frac {0+5+y_3}{3}\)

y

_{centroid}= 1

1 = \(\frac {0 + 5 + y_3}{3}\)

3 = 5 + y

_{3}

-2 = y

_{3}

The third coordinate is (5, -2).

**Sanfoundry Global Education & Learning Series – Mathematics – Class 10**.

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