Question

Find the point on x-axis which is equidistant from (2,-5) and (-2,7)

Answer 1

Answer:

The point on x-axis is (x,0)

(x-2)^2 +25=(x+2)^2 +49

-8x=24

x=-3

Therefore the point on the x-axis is (-3,0)

Answer 2

GIVEN:-

Two points are A(2,-5) and B(-2,7)

TO FIND:-

A point on x axis that is equidistant from the given two points

EXPLANATION:-

As the point is on x-axis ,y-coordinate is 0

let the point that is equidistant from A and B be P(x,0)

$\underline{\sf Distance \:b/w\: two\: points =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

Finding PA:-

$PA=\sqrt{(2-x)^2+(-5)^2}$

$\longrightarrow PA=\sqrt{(2-x)^2+25}$

Finding PB:-

$PB=\sqrt{(x+2)^2+(7)^2}$

$\longrightarrow PB=\sqrt{(x+2)^2+49}$

According to the given information

PA=PB

squaring on both sides

PA²=PB²

${(2 - x)}^{2} + 25 = {(x + 2)}^{2} + 49$

expanding using

(a+b)²=a²+b²+2ab

$= > 4 + {x}^{2} - 4x = {x }^{2} + 4 + 4x + 24 \\ = > 4 - 4 + {x}^{2} - {x}^{2} - 8x = 24 \\ x = \dfrac{24}{ - 8} \\ \boxed{ \sf \: x = - 3}$

So the point is (-3,0)