Let A (5, 9) and B (-4, 6) be the given points
Let the point on x – axis equidistant from the above points be C(x, 0)
Now, we have
AC = √[(x – 5)2 + (0 – 9)2]
= √[x2 – 10x + 25 + 81]
= √[x2 – 10x + 106]
And,
BC = √[(x – (-4))2 + (0 – 6)2]
= √[x2 + 8x + 16 + 36]
= √[x2 + 8x + 52]
As AC = BC (given condition)
So, AC2 = BC2
x2 – 10x + 106 = x2 + 8x + 52
18x = 54
x = 3
Therefore, the point on the x-axis is (3, 0)