What is the point on x-axis is equidistant from (7,6) (-3,4) .
Let the point on x-axis be P (x, 0).
The given points are A(7, 6) and B(–3, 4).
Since, point P is equidistant from A and B, therefore AP = BP
or AP2 = BP2
⇒ (7 – x)2 + (6 – 0)2 = (–3 – x)2 + (4 – 0)2
⇒ 49 + x2 – 14x + 36 = 9 + x2 + 6x + 16
⇒ 85 – 14x = 6x + 25
⇒ 20x = 60
⇒ x = 3
Hence, the required point is (3, 0).