Question

find the point on x-axis which is equidistant from p (2,-5) and q (-2,9)​

Answer 1

Answer:

that you will find this your book

Answer 2

Answer:

The answer is -7

Step-by-step explanation:

Concept : Distance formula is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let required point on x-axis is $A(x,0)$

Given  $\text{AP}=\text{AQ}$

 $\Rightarrow \sqrt{(x-2)^2+(0+5)^2}=\sqrt{(x+2)^2+(0-9)^2}$

Square both sides

$\Rightarrow x^2-4x+4+25=x^2+4x+4+81\\\Rightarrow -4x+25=4x+81\\\Rightarrow 4x+4x=25-81\\\Rightarrow 8x=-56\\\Rightarrow x=-\dfrac{56}{8}=-7$