RS Aggarwal Veena Aggarwal Class 9 Chapter 5 Solution

Get solution of all questions of RS Aggarwal Veena Aggarwal Class 9 Chapter 5 (Coordinate Geometry).

RS Aggarwal and Veena Aggarwal Class 9 Chapter 5 Exercise 5 Solution

1: On the plane of a graph paper draw X’OX and YOY’ as a coordinate axes and plot each of the following point.

(i) A(5, 3)
(ii) B(6, 2)
(iii) C(-5, 3)
(iv) D (4, -6)
(v) E(-3, -2)
(vi) F(-4, 4)
(vii) G(3, -4)
(viii) H (5, 0)
(ix) I(0, 6)
(x) J (-3, 0)
(xi) K(0, -2)
(xii) O(0, 0)

Solution:

2: Write down the coordinates of each of the following points A, B, C, D and E.

Coordinates of A (-6, 5), B(5, 4), C(-3, 2), D(2, -2) and E(-1, -4)

3: For each of the following points, write the quadrant in which it lies.

(i) (-6, 3)
Lies in second quadrant
(ii) (-5, -3)
Lies in third quadrant
(iii) (11, 6)
Lies in first quadrant
(iv) (1, -4)
Lies in fourth quadrant
(v) (-7, -4)
Lies in third quadrant
(vi) (4, -1)
Lies in fourth quadrant
(vii)(-3, 8)
Lies in second quadrant
(viii) (3, -8)
Lies in fourth quadrant

4: Write the axis on which the given point lies.

(i) (2, 0)
Lies on x – axis
(ii) (0, -5)
Lies on y – axis
(iii) (-4, 0)
Lies on x – axis
(iv) (0, -1)
Lies on y – axis

5: Which of the following points lies on the x- axis ?

(i) A(0 , 8)
(ii) B(4, 0)
(iii) C(0, -3)
(iv) D(-6, 0)
(v) E(2, 1)
(vi) F(-2, -1)
(vii) G(-1, 0)
(viii) H(0, -2)

Points lies on x – axis,
(ii) B(4, 0) , (iv) D(-6, 0) and (vii) G(-1, 0)

6: Plot the points A(2, 5) , B(-2, 2) and C(4, 2) on a graph paper. Join AB, BC and AC. Calculate the area of triangle ABC.

Solution

From figure,
BC = BM + CM
= 4 + 2 = 6
AM = AN – MN
= 5 – 2 = 3
∴ Area of triangle ABC = 1/2 (AM×BC)=1/2 (3×6) = 9

Hence the area of triangle is 9 sq. unit.

7: Three vertices of a rectangle ABCD are A (3, 1), B(-3, 1) and C(-3, 3). Plot these points on the graph paper and find the coordinates of the fourth vertex D. Also, find the area of rectangle ABCD.

Since we have given ABCD is a rectangle,

So point D will be image of C therefore Point D will be (3, 3)

Now, BC = CM – BM = 3 – 1 = 2
And CD = CE + ED = 3+ 3 = 6

∴ Area of rectangle ABCD = CD×BC
=6×2=12 sq. unit

Hence area of rectangle ABCD is 12 sq. unit.