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

Dear Student,

Please find below the solution to the asked query:

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 ) =  $\sqrt{{\left(x_{ 2} - x_{ 1}\right)}^{ 2}+ {\left(y_{ 2} - y_{ 1}\right)}^{ 2}}$

For AX , x${}_1$ = - 2 ,  x${}_2$ = x  And ​ y${}_1$ = 5 ,  y${}_2$ = 0 Substitute all values in distance formula we get

AX = $\sqrt{{\left(x_{ } - \left( -2 \right)\right)}^{ 2}+ {\left( 0 - 5\right)}^{ 2}}$

AX² = ( x  + 2 )² + ( - 5 )²

AX² = x² + 4 + 4 x  + 25

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

And

For BX , x${}_1$ = 2 ,  x${}_2$ = x  And ​ y${}_1$ = - 3 ,  y${}_2$ = 0 Substitute all values in distance formula we get

BX = $\sqrt{{\left(x_{ } -2\right)}^{ 2}+ {\left( 0 -\left(-3\right)\right)}^{ 2}}$

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 )


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