Question

Find the coordinates of the point on x-axis which is equidistant from the points ( - 2, 5 ) and ( 2, - 3 ).

Answer 1

Answer:

(-2 , 0)

Step-by-step explanation:

Let, Given points be A (-2 , 5) and B (2 , -3)

Let, Required poin be P (x , y)

Here, it is given point lies on "X axis" then Y co-ordinate must be 0.

Then,

ATQ, PA = PB

Using Distance Formula:- D = √(x₁ - x₂)² + (y₁ - y₂)²

=> √[x-(-2)]² + 5²   =    √(x-2)² +  (-3)²

=> x² + 4 + 4x + 25  =  x² + 4 -4x + 9

=> 8x = -16

=> x = -2

Thus, co-ordinate of P (-2,0)

_______________

Amrit_____

Answer 2

Hey there !! 


Let the given points be A( -2 , 5 ) and B( 2 , -3 ) .

And, let the required point on x-axis be P( x , 0 ) .

Then,

→ PA = PB . [ A and B are equidistant from P ] 

[ Squaring both side ] 

=> PA² = PB² .

[ Using distance formula ] 

$\begin{lgathered}= > {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} = {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} . \\ \\\end{lgathered}$

=> [ x - ( -2 ) ]² + ( x - 5 )² = ( x - 2 )² + [ 0 - ( -3 ) ]² .

=> ( x + 2 )² + ( 0 - 5 )² = ( x - 2 )² + ( 0 + 3 )² .

=> ( x + 2 )² + 25 = ( x - 2 )² + 9 .

=> ( x + 2 )² - ( x - 2 )² = 9 - 25 .

=> [ ( x + 2 ) + ( x - 2 ) ] [ ( x + 2 ) - ( x - 2 ) ] = - 16 .

=> ( x + 2 + x - 2 ) ( x + 2 - x + 2 ) = - 16 .

=> ( 2x ) ( 4 ) = - 16 .

=> 8x = - 16 .

=> x = -16/8 .

•°• x = -2 .

✔✔ Hence, the required point on x-axis is ( -2 , 0 ) ✅✅ .


THANKS


#BeBrainly.