Find the point on the X-axis which is equidistant from (2,-5) and (-2,9)
Answers
GIVEN :
Two points (2,-5) and (-2,9)
TO FIND :
The point which is equidistance from the two given points on which is lieing on the X - axis
SOLUTION :
let the point be P which is equidistance from the given points
P = (x,y)
let the given two points be A and B
The distance between the two points P. and A is
PA^2 = (x-2)^2. + (y+5)^2
PA^2 = x^2. + y^2 - 4x + 10y +. 29
The distance between the two points P and B is
PB^2 = (x+2)^2 + (y-9)^2
PB^2 = x^2 +y^2 + 4x - 18y + 85
Given That Two Points Are Equidistance so
PA^2 = PB^2
x^2 + y^2 - 4x + 10y + 29 = x^2 +. y^2 + 4x - 18y + 85
- 8x + 28y - 56 = 0
8x - 28y + 56 = 0
2x - 7y + 14 = 0
2x - 7y = - 14
- x/7 + y/2 = 0
THEREFORE THE POINTS ARE
. (-7,2)
And the point on x-axis is -7