# Coordinate Geometry

## NCERT Exercise

### 7.2 Part 2

Question 5: Find the ratio in which the line segment joining A (1, -5) and B (-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Solution: We have; x_1 = 1, y_1 = 5, x_2 = -4, y_2 = 5

Coordinates of a point on x-axis = (x, 0)

Let us assume the ratio = k : 1

x=(1-4k)/(k+1)-------(1)

0=(-5+5k)/(k+1)

Or, -5+5k=0
Or, 5k=5
Or, k=1

Substituting this value in equation (1), we get;

x=(1-4)/(1+1)=-3/2

Ratio = (1 : 1)

Point C = (-3/2, 0)

Question 6: If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Solution: A = (1, 2), B = (4, y), C = (x, 6), D = (3, 5)

AC = BD (diagonals of parallelogram are equal)

We know that diagonals of a parallelogram bisect each other. Let us assume point O is the point of intersection of two diagonals. Then coordinates of point O can be given as follows:

For diagonal AC;

O=((1+x)/(2)\, (2+6)/(2))

=((1+x)/(2)\, 4)--------(1)

For diagonal BD;

O=((3+4)/(2)\, (5+y)/(2))

=((7)/(2)\, (5+y)/(2))-------(2)

From equations (1) and 2;

(1+x)/(2)=7/2

Or, 1+x=7
Or, x=6
(5+y)/(2)=4

Or, 5+y=8
Or, y=3

Hence, x = 6, y = 3

Question 7: Find the coordinates of a point A, where AB is the diameter of a circle, whose centre is (2, -3) and B is (1, 4).

Solution: Centre O bisects AB and AB = (x, y). Hence;

2=(x+1)/(2)

Or, x+1=4
Or, x=3

Similarly,

-3=(y+4)/(2)

Or, y+4=-6
Or, y=-10

Hence; O = (3, -10)

Question 8: If A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = 3/7(AB) and P lies on the line segment AB.

Solution:

AP=3/7\AB

Or, (AP)/(AB)=3/7

Or, (AP)/(PB)=3/4

This means that point P divides AB in 3 : 4 ratio. Coordinates of P can be calculated as follows:

x=(4xx(-2)+3xx2)/(7)

=(-8+6)/(7)=-2/7

y=(4xx(-2)+3xx(-4))/(7)

=(-8-12)/(7)=-20/7

Hence, P = (-2/7, -20/7)

Question 9: Find the coordinates of the points which divide the line segment joining A (-2, 2) and B (2, 8) into four equal parts.

Solution: Let us assume point Q bisects AB; as shown in figure.

Coordinates of Q can be calculated as follows:

x=(2-2)/(2)=0

y=(2+8)/(2)=5

Or, Q=(0,5)

Now, point P bisects AQ.

Coordinates of P can be calculated as follows:

x=(-2+0)/(2)=-1

y=(2+5)/(2)=7/2

Or, P=(-1,\7/2)

Point R bisects QB.

Coordinates of R can be calculated as follows:

x=(0+2)/2=1

y=(5+8)/(2)=13/2

Or, R(1,\13/2)

Hence, P = (-1, 7/2), Q = (0, 5) and R = (1, 13/2)

Question 10: Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.

Solution: A = (3, 0), B = (4, 5), C = (-1, 4), D = (-2, -1)

Diagonal AC can be calculated as follows by using distance formula:

AC=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

=sqrt((-1-3)^2+(4-0)^2)

=sqrt((-4)^2+4^2)

=sqrt(16+16)=4sqrt2

Similarly, diagonal BD can be calculated as follows:

BD=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

=sqrt((-2-4)^2+(-1-5)^2)

=sqrt((-6)^2+(-6)^2)

=sqrt(36+36)=6sqrt2

Area of rhombus can now be calculated as follows:

1/2\xxd_1xxd_2

=1/2\xx4sqrt2xx6sqrt2=24 sq unit