# Coordinate Geometry

## NCERT Exercise

### 7.1 Part 5

Question 7: Find the point on the x-axis which is equidistant from (2, –5) and (– 2, 9).

Solution: Let given coordinates are A(2, –5), BC (–2, 9) and P(x, 0) is on the x-axis.

Since, point P is equidistanct from the given coordinates,

thus AP = BP

AP=sqrt((x_2-x_2)^2+(y_2-y_1)^2)

Here, x1 = 2, x2 = x, y1 = -5 and y2 = 0

So, AP=sqrt((x-2)^2+(0+5)^2)

=sqrt(x^2+4-4x+25)

=sqrt(x^2-4x+29)

BP= sqrt((x_2-x_2)^2+(y_2-y_1)^2)

Here, x1 = -2, x2 = x, y1 = 9 and y2 = 0

So, BP=sqrt((x+2)^2+(0-9)^2)

=sqrt(x^2+4+4x+81)

=sqrt(x^2+4x+85)

Now, since AP=BP

So, sqrt(x^2-4x+29)=sqrt(x^2+4x+85)

After squaring both sides we get:

x^2-4x+29=x^2+4x+85

Or, -4x+29=4x+85

Or, -8x+29=85

Or, -8x=85-29=56

Or, -x=56/8=7

Or, x=-7

Thus, required point is (-7, 0)

Question 8: Find the value of y for the distance between the points P(2, – 3) and Q (10, y) is 10 units.

Solution: Given, P(2, –3) and Q(10, y) and QP = 10 unit

PQ=sqrt((x_2-x_2)^2+(y_2-y_1)^2)

Here, x1 = 2, x2 = 10, y1 = -3 and y2 = y

So, PQ=sqrt((10-2)^2+(y+3)^2)
Or 10=sqrt(8^2+(y+3)^2)
Or 10=sqrt(64+y^2+9+6y)

After squaring both sides we get;

10^2=64+9+y^2+6y
Or, 100=73+y^2+6y
Or, y^2+6y=100-73=27
Or, y^2+6y-27=0
Or, y^2+9y-3y-27=0
Or, y(y+9)-3(y+9)=0
Or, (y-3)(y+9)=0

Thus, if y-3=0 Then y=3
And if y+9=0
Then y=-9

Thus, the required value = 3 or -9

Thus, required value = 3 or – 9