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) = √[(x2-x1)2+(y2-y1)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) = √[(x2-x1)2+(y2-y1)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
x2+6x+9+16 = x2-2x+1+16
x2+6x+25 = x2-2x+17
8x = -8
x = -8/8 = -1
Hence the point on X axis which is equidistant from A and B is C(-1,0)
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