Let C be the point on X axis equidistant from A(-3,4) and B(1,-4).

Since C lies on X axis, the Y co-ordinate of C is 0.

Let C = (x,0)

Co-ordinates of A = (-3, 4)

Co-ordinates of B = (1, -4)

Since C is equidistant from A and B ,

AC = BC

By distance formula, d(A,C) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

d(A,C) = √[(x-(-3))^{2}+(0-4)^{2}]

d(A,C) = √[(x+3)^{2}+(4)^{2}]

d(A,C) = √[(x+3)^{2}+16] ……(i)

By distance formula, d(B,C) = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

d(B,C) = √[(x-1)^{2}+(0-(-4))^{2}]

d(B,C) = √[(x-1)^{2}+(4)^{2}]

d(B,C) = √[(x-1)^{2}+16] ……(ii)

Equating (i) and (ii) [∵AC = BC]

√[(x+3)^{2}+16] = √[(x-1)^{2}+16]

Squaring both sides

(x+3)^{2}+16 = (x-1)^{2}+16

x^{2}+6x+9+16 = x^{2}-2x+1+16

x^{2}+6x+25 = x^{2}-2x+17

8x = -8

x = -8/8 = -1

Hence the point on X axis which is equidistant from A and B is C(-1,0)