Question

Find the co-ordinates of a points on x-axis which is equidistant from the points (–2,5) and (2,–3).

Answer 1

Coordinate geometry is the branch of mathematics which deals with the position of an object lying in a plane.

Each  point in cartesian plane has two coordinates X coordinate and Y coordinate.

The X co-ordinate is called the abscissa.

The Y co-ordinate is called the ordinate.

Coordinates X and Y taken together are called coordinates of a point. (x,y) is called an ordered pair.

DISTANCE FORMULA:

The distance between any two points A(x1,y1) and B(x2,y2) is:

AB=√(x2 - x1)² + (y2 - y1)²]

SOLUTION:

The coordinates of every point on the x- axis are of the form (x,0)

Given : A( - 2 ,5) and B( 2 , - 3 ) are equidistant from P( x,0), AP  = BP .
For distance AP : x1 = - 2 ,y1 = 5 , x2 = x & ​y2 = 0
Distance of AP =√ (x− (−2)² + (0 − 5)²

[Distance Formula=√(x2 - x1)² + (y2 - y1)²]
AP= √( x  + 2 )² + (- 5 )²
= √x² + 2² +2×2×x + 5²
[(a+b)² = a² + b² +2ab]
AP= √x² +4 +4x +25
AP = √x² + 4 x  + 29…………………(1)                

For distance BP : x1 = 2 , x2 = x & ​ y1 = - 3 , y2 = 0
BP =√ (x −2) ² + ( 0 −(−3))²

[Distance Formula=√(x2 - x1)² + (y2 - y1)²]

BP = √(x - 2 )² + ( 3 )²
BP= √x² + 4 - 4 x  + 9
[(a- b)² = a² + b² - 2ab]
BP= √x² - 4 x  + 13…………………..(2)                 

AP = BP [equidistant from P]
√x² + 4 x  + 29 = √x² - 4 x  + 13
[From eq. 1 and 2 ]

On squaring both sides
x² + 4 x  + 29 = x² - 4 x  + 13
x² - x²  + 4 x + 4 x=  + 13 - 29
8 x = - 16
x = -16/8
x = - 2
Hence , the point on x - axis is (-2 , 0) .   

HOPE THIS WILL HELP YOU...

Answer 2

y- y1 devided by x- x1 is equal to y1- y devided by x1- x2 use this law