# Mathematics Questions and Answers – Geometry – Section Formula

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

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)

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)

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 = $$\frac {5}{2}$$

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)

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)

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)

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

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

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)

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

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)

Explanation: We know, xcentroid = $$\frac {x_1+x_2+x_3}{3}$$ and ycentroid = $$\frac {y_1+y_2+y_3}{3}$$
xcentroid = $$\frac {-2+0+4}{3} = \frac {2}{3}$$
ycentroid = $$\frac {4+0+2}{3}$$ = 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)

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)

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)

Explanation: The two vertices of triangle are A (-3, 0) and B (-8, 5). Its centroid is (-2, 1).
We know, xcentroid = $$\frac {x_1+x_2+x_3}{3}$$ and ycentroid = $$\frac {y_1+y_2+y_3}{3}$$
Now, xcentroid = $$\frac {-3-8+x_3}{3}$$
xcentroid = -2
-2 = $$\frac {-3 – 8 + x_3}{3}$$
-6 = -3 – 8 + x3
5 = x3
Now, ycentroid = $$\frac {0+5+y_3}{3}$$
ycentroid = 1
1 = $$\frac {0 + 5 + y_3}{3}$$
3 = 5 + y3
-2 = y3
The third coordinate is (5, -2).

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