Class 10 Section Formula — practice questions with answers

Geometry — Coordinates · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.

1. The midpoint of the line segment joining A(2, 4) and B(6, 8) is:

A(4, 6)
B(8, 12)
C(2, 2)
D(4, 4)

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Answer: A) (4, 6)

Midpoint = (

\frac{2 + 6}{2}

\frac{4 + 8}{2}

) = (

\frac{8}{2}

\frac{12}{2}

) = (4, 6).

2. The midpoint of segment joining (7, -2) and (-3, 6) lies in which quadrant?

AFirst quadrant
BSecond quadrant
CThird quadrant
DFourth quadrant

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Answer: A) First quadrant

Midpoint = (

\frac{7 - 3}{2}

\frac{- 2 + 6}{2}

) = (2, 2), both positive => first quadrant.

3. Points A(1, -1), B(5, 3) and C are such that B divides AC in ratio 2:3 (A to C).
Find the coordinates of C.

A(11, 9)
B(9, 7)
C(7, 5)
D(13, 11)

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Answer: A) (11, 9)

B=(

\frac{2 \cdot C x + 3 \cdot 1}{5}

\frac{2 \cdot C y + 3 \cdot ( - 1 )}{5}

)=(5,3). 2Cx+3=25 => Cx=11; 2Cy-3=15 => Cy=9. C=(11, 9).

4. A water pipeline runs straight from a pumping station P(-5, 2) to a village tank T(10, 12) on a survey grid marked in kilometres.
A pressure valve is fitted at the point dividing PT in the ratio 2 : 3 measured from P.
Where is the valve?

A(1, 6)
B(4, 8)
C(2.5, 7)
D(1, 5)

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Answer: A) (1, 6)

x =

\frac{2 \cdot 10 + 3 \cdot ( - 5 )}{5}

\frac{20 - 15}{5}

= 1 and y =

\frac{2 \cdot 12 + 3 \cdot 2}{5}

\frac{24 + 6}{5}

= 6, so the valve is at (1, 6). (4, 8) applies the ratio as 3 : 2, i.e. measured from T instead of from P; (2.5, 7) is the midpoint, which ignores the 2 : 3 split; (1, 5) divides 30 by 6 instead of by m + n = 5.

5. A, B and C are three points on a straight road, with A(1, -2) and B(9, 6), and B divides AC in the ratio 4 : 3.
A milestone M is set at the point dividing AB in the ratio 1 : 3.
Find C and M.

AC(13, 10) and M(3, 0)
BC(15, 12) and M(5, 2)
CC(15, 12) and M(7, 4)
DC(15, 12) and M(3, 0)

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Answer: D) C(15, 12) and M(3, 0)

For C:

\frac{4 x + 3 \cdot 1}{7}

= 9 gives 4x = 60 and x = 15;

\frac{4 y + 3 \cdot ( - 2 )}{7}

= 6 gives 4y = 48 and y = 12, so C = (15, 12). For M on AB with ratio 1 : 3: x =

\frac{1 \cdot 9 + 3 \cdot 1}{4}

= 3 and y =

\frac{1 \cdot 6 + 3 \cdot ( - 2 )}{4}

= 0, so M = (3, 0). C(13, 10) mis-solves 4x + 3 = 63 as 4x = 52; M(5, 2) treats M as the midpoint of AB; M(7, 4) uses the ratio 3 : 1 instead of 1 : 3.

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