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