Question

find the point x-axis which is equidistant from points (-2,5) and (2,-3)

Answer 1

hii dear here is your answer......
according to question...
Let the point of x - axis be X (x ,  0 )

Given : A( - 2 , 5) and B( 2 , - 3 ) are equidistant from P , So

AX  = BX , Then

AX2 =  BX2                                                                ---- ( A ) 

we can use distance formula , That is Distance ( d ) =  (x 2 − x 1) 2+ (y 2 − y 1) 2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

For AX , x1 = - 2 ,  x2 = x  And ​ y1 = 5 ,  y2 = 0 Substitute all values in distance formula we get

AX = (x   − ( −2 )) 2+ ( 0 − 5) 2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

AX2 = ( x  + 2 )2 + ( - 5 )2

AX2 = x2 + 4 + 4 x  + 25

AX2 = x2 + 4 x  + 29                                                     ---- ( 1 )

And

For BX , x1 = 2 ,  x2 = x  And ​ y1 = - 3 ,  y2 = 0 Substitute all values in distance formula we get

BX = (x   −2) 2+ ( 0 −(−3)) 2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

BX2 = ( x  - 2 )2 + ( 3 )2

BX2 = x2 + 4 - 4 x  + 9

BX2 = x2 - 4 x  + 13                                                     ---- ( 2 )

From equation A , 1 and 2 we get 
x2 + 4 x  + 29   = x2 - 4 x  + 13       

8 x  = - 16

x  = - 2

Hence the point on x - axis is (-2 , 0)                                                   ( Ans )

hope it helps you.....

Answer 2

Answer:

This was done using distance formula.