Class 10 Distance 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. What is the distance formula between points A(x1, y1) and B(x2, y2)?

A $(x 2 - x 1)^{2} + (y 2 - y 1)^{2}$
B(x2 - x1)² + (y2 - y1)²
C $(x 2 + x 1)^{2} + (y 2 + y 1)^{2}$
D|x2 - x1| + |y2 - y1|

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Answer: A)

(x 2 - x 1)^{2} + (y 2 - y 1)^{2}

The distance formula is

(x 2 - x 1)^{2} + (y 2 - y 1)^{2}

, derived from the Pythagoras theorem.

2. The distance between (-2, 4) and (3, 4) is:

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Answer: A) 5

Same y-coordinate, so distance = |3 - (-2)| = 5.

3. Show that the points A(1, 7), B(4, 2), C(-1, -1) and D(-4, 4) taken in order form a square by checking sides and diagonals. What is the length of each side?

A $34$
B $50$
C6
D $26$

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Answer: A)

34

AB =

(4 - 1)^{2} + (2 - 7)^{2} = 9 + 25 = 34

. Similarly BC = CD = DA =

34

, and diagonals AC = BD =

68

, so it is a square with side

34

4. A flower bed has corners at A(-1, 1), B(2, 5), C(7, 5) and D(4, 1), the units being metres. Metal edging is to be fixed all the way round the bed.
What length of edging is needed?

A16 m
B20 m
C24 m
D25 m

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Answer: B) 20 m

A B = (2 + 1)^{2} + (5 - 1)^{2} = 9 + 16 = 5

, BC = |7 - 2| = 5,

C D = (4 - 7)^{2} + (1 - 5)^{2} = 5

and DA = |4 - (-1)| = 5. All four sides measure 5 m, so the edging needed is

4 \times 5 = 20

m. The bed is a rhombus, not a square: its diagonals are

A C = 64 + 16 = 45

and

B D = 4 + 16 = 25

, which are unequal. 25 m comes from squaring the side instead of adding the four sides.

5. Three cottages stand at P(-4, 0), Q(0, 8) and R(4, 0) on a plan marked in metres. A well is to be dug at the one spot that is the same distance from all three cottages.
How far will the well be from each cottage?

A8 m
B $45$ m
C5 m
D4 m

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Answer: C) 5 m

P and R are mirror images in the y-axis, so the well must lie on the y-axis, say at (0, k). Setting its distance to P equal to its distance to Q gives

16 + k^{2} = ∣8 - k ∣

, so

16 + k^{2} = 64 - 16 k + k^{2}

, which leaves 16k = 48 and k = 3. The well is at (0, 3), and

(- 4 - 0)^{2} + (0 - 3)^{2} = 16 + 9 = 5

m to P, the same 5 m to R, and |8 - 3| = 5 m to Q. The value 4 m is only half of PR, and

45

is the side PQ.

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