Question

Find a point on the x axis equidistant from the point (2,-5)and(-2,9)

Answer 1

Question :

Find a point on the x axis equidistant from the point (2,-5)and(-2,9).

Solution :

Let a point on the x-axis be P (h,0)

and point A(2,-5),B ( -2,9)

According to the Question

Point P is equidistant from A and B

➝BP=AP

By distance Formula

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

On squaring both sides

$\implies\sf\:x^2+4+4x+81=x^2+4-4x+25$

$\implies\sf\:8x=-56$

$\implies\sf\:x=\dfrac{-56}{8}=-7$

Thus , (h,0)=(-7,0)

Hence the point is (-7,0) which is equidistant from (2,-5) and (-2,9).

Answer 2

$\red{\huge{\boxed{\sf{Given:-}}}}$

Two point (2,-5)and(-2,9)

$\red{\huge{\boxed{\sf{To\ be\ found:-}}}}$

A point on the x-axis equidistant from the point (2,-5)and(-2,9)

So,

Let the point be P(x,y)

but we have to find point on the x-axis so, points will be (x,0)

Now,

Using The Distance formula

(PA)² = (PB)²

→ PA = PB

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

$\implies \bf x^{2} + 4-4x+ 25 =x^{2} +4x+4+81$

$\implies \bf \cancel{x^{2}} \cancel{+4} - 4x + 25 = \cancel{x^{2}} +4x \cancel{+4} +81$

$\implies \bf- 4x + 25 = 4x +81$

$\implies \bf- 4x-4x = 81 -25$

$\implies \bf-8 x = 56$

$\implies \bf x = \dfrac{56}{-8}$

$\implies \boxed{\bf x = -7 }$

Hence,

The point P is (-7, 0) that is equidistant from the point A(2,-5) and B(-2,9)