Coordinate Geometry — Chapter Test | Class 10 IMO

Q1

The distance of any point from itself is:

A point coincides with itself, so the distance is 0.

Q2

The distance between (7, 1) and (3, 4) is:

√((7 − 3)² + (1 − 4)²) = √(16 + 9) = 5.

Q3

The distance between the points (0, 0) and (3, 4) is:

Distance = √(3² + 4²) = √25 = 5.

Q4

The area of the quadrilateral with vertices (1, 1), (7, 1), (7, 5) and (1, 5) is:

It is a 6 by 4 rectangle, so the area is 6 × 4 = 24 sq units.

Q5

Three sensors are placed at (0, 0), (4, 0) and (2, 6). The area they enclose is:

Base 4 (along the x-axis) and height 6, so area = ½ × 4 × 6 = 12.

Q6

The distance of the point (5, 12) from the origin is:

√(5² + 12²) = √169 = 13.

Q7

The point (−3, 5) lies in which quadrant?

Negative x with positive y places it in Quadrant II.

Q8

The point that divides the join of (1, 2) and (4, 8) in the ratio 1 : 2 internally is:

Using the section formula: ((1·4 + 2·1)/3, (1·8 + 2·2)/3) = (2, 4).

Q9

A surveyor marks a plot with corners (0, 0), (4, 0), (4, 3) and (0, 3). Its perimeter is:

It is a 4 by 3 rectangle, so perimeter = 2(4 + 3) = 14 units.

Q10

The reflection of (3, 5) in the y-axis is:

Reflecting in the y-axis changes the sign of the x-coordinate.

Q11

The midpoint of the segment joining (2, 4) and (6, 8) is:

Midpoint = ((2 + 6)/2, (4 + 8)/2) = (4, 6).

Q12

A garden is a triangle with corners at (0, 0), (6, 0) and (6, 8) on a grid (metres). Its area is:

Right-angled with legs 6 and 8, so area = ½ × 6 × 8 = 24 m².

Q13

The area of the triangle with vertices (1, 1), (2, 3) and (4, 5) is:

Area = ½|1(3 − 5) + 2(5 − 1) + 4(1 − 3)| = ½|−2 + 8 − 8| = 1.

Q14

The points (3, 0), (6, 4) and (−1, 3) are the vertices of:

The side lengths are 5, 5 and 5√2, and 5² + 5² = (5√2)², so it is a right isosceles triangle.

Q15

The reflection of the point (3, 5) in the x-axis is:

Reflecting in the x-axis changes the sign of the y-coordinate.