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

Explanation

`P(2,-5)` and `Q(-2,9)`
Let point `A(x,y)` on X-axis be equidistant from the points P and Q.
`:.` its y coordinate is 0.
`:.A(x,0)`
Now `AP=AQ`. ……..(Given)
`:.` by distance formula,
`sqrt((x-2)^(2)+[0-(-5)]^(2))=sqrt([x-(-2)]^(2)+(0-9)^(2))`
`:.sqrt((x-2)^(2)+5^(2))=sqrt((x+2)^(2)+(-9)^(2))`
`:.sqrt((x-2)^(2)+25)=sqrt((x+2)^(2)+81)`
Squaring both the sides we get
`(x-2)^(2)+25=(x+2)^(2)+81`
`:.x^(2)-4x+4+25=x^(2)+4x+4+81`
`:.-4x-4x=81-25`
`:.-8x=56`
`:.x=56/(-8)`
`:.x=-7`
`:.` the coordinates of point A are `(-7,0)`.
`(-7,0)` lies on X-axis and is equidistant from the points `P(2,-5)` ankd `Q(-2,9)`.