Find the point on the X-axis which is equidistant from (-3,4)
and B(1,-4).

Explanation

`A(-3,4)` and `B(1,-4)`.
Let P the point on X-axis equidistant from point A and B.
`:.PA=PB`………1
P lies on X-axis `:.` its y coordinate is 0.
Let `P(x,0)`
Now `PA=PB`…..[From 1]
By distance formula,
`sqrt([x-(-3)]^(2)+(0-4)^(2))=sqrt((x-1)^(2)+[0-(-4)]^(2))`
`:.sqrt((x+3)^(2)+(-4)^(2))=sqrt((x-1)^(2)+(40)^(2))`
Squaring both sides
`(x+3)^(2)+16=(x-1)^(2)+16`
`:.x^(2)+6x+9=x^(2)-2x+1`
`:.6x+2x=1-9`
`:.8x=-8`
`:.x=-8/8`
`:.x=-1`
`:.P(-1,0)`
The coordinates of point on X-axis equidistant from points A and B are `(-1,0)`.