Co-ordinate Geometry — Exercise 21(A)

Write down the co-ordinates of each of the following points A, B, C, D, E shown below on the graph paper.

Answer

A(-3, 2)

B(2, 1)

C(4, -2)

D(-1, -3)

E(0, -1)

Plot each of the following points on a graph paper.

(i) A(6, 3)

(ii) B(-4, 1)

(iii) C(-2, -5)

(iv) D(2, -5)

(v) P(4, 0)

(vi) Q(0, 3)

(vii) R(-3, -3)

(viii) S(0, -3)

Answer

On which axis does the following point lie?

(i) (5, 0)

(ii) (0, -2)

(iii) (0, 3)

(iv) (-3, 0)

Answer

(i) (5, 0)

The point whose y-coordinate = 0, lies on the x-axis.

Hence, point (5, 0) lies on x-axis.

(ii) (0, -2)

The point whose x-coordinate = 0, lies on the y-axis.

Hence, point (0, -2) lies on y-axis.

(iii) (0, 3)

The point whose x-coordinate = 0, lies on the y-axis.

Hence, point (0, 3) lies on y-axis.

(iv) (-3, 0)

The point whose y-coordinate = 0, lies on the x-axis.

Hence, point (-3, 0) lies on x-axis.

In which quadrant does the given point lie?

(i) A(-3, 2)

(ii) B(-5, -3)

(iii) C(2, -7)

(iv) D(-2, -2)

Answer

(i) A(-3, 2)

Here, x is negative and y is positive.

Hence, point A lies in the 2nd quadrant.

(ii) B(-5, -3)

As x is negative, y is negative

Hence, point B lies in the 3rd quadrant.

(iii) C(2, -7)

As x is positive and y is negative

Hence, point C lies in the 4th quadrant.

(iv) D(-2, -2)

As x is negative, y is negative

Hence, point D lies in the 3rd quadrant.

The points A(2, -2), B(8, 4) and C(5, 7) are three vertices of a rectangle ABCD. Plot these points on a graph paper and hence, find the co-ordinates of its fourth vertex D.

Answer

Given,

A(2, -2), B(8, 4) and C(5, 7) are three vertices of a rectangle ABCD.

As we know diagonals of a rectangle are equal and bisect each other.

Therefore,

Midpoint of AC = Midpoint of BD

Midpoint of AC = $\Big(\dfrac{2 + 5}{2}, \dfrac{-2 + 7}{2}\Big)$

= $\Big(\dfrac{7}{2}, \dfrac{5}{2}\Big)$

Let D = (x, y)

Midpoint of BD:

$\Big(\dfrac{8 + x}{2}, \dfrac{4 + y}{2}\Big)$

Now Equating both midpoints

$\dfrac{8 + x}{2} = \dfrac{7}{2}$

$\dfrac{4 + y}{2} = \dfrac{5}{2}$

Solving for x,

$\dfrac{8 + x}{2} = \dfrac{7}{2}$

8 + x = 7

x = 7 - 8

x = -1

Solving for y,

$\dfrac{4 + y}{2} = \dfrac{5}{2}$

4 + y = 5

y = 1.

Hence, coordinates of D = (-1, 1).

The points A(3, 2), B(0, 5) and D(0, -1) are the three vertices of a square ABCD. Plot these points on a graph paper and hence find the co-ordinates of the vertex C.

Answer

A(3, 2), B(0, 5) and D(0, -1) are the three vertices of a square ABCD.

As we know diagonals of square are equal and bisect each other.

Therefore,

Midpoint of AC = Midpoint of BD

Let C = (x, y)

Midpoint of AC = $\Big(\dfrac{3 + x}{2}, \dfrac{2 + y}{2}\Big)$

Midpoint of BD :

$\Big(\dfrac{0 + 0}{2}, \dfrac{5 + (-1)}{2}\Big)$

(0, 2)

Now Equating both midpoints,

$\dfrac{3 + x}{2}$ = 0

$\dfrac{2 + y}{2}$ = 2

Solving for x,

$\dfrac{3 + x}{2}$ = 0

3 + x = 0

x = -3

Solving for y,

$\dfrac{2 + y}{2}$ = 2

2 + y = 4

y = 2.

Hence, coordinates of C = (-3, 2).