Given:
Two points are A(7, 6) and B(3, 4).
Let the required point on the x-axis be P(x, 0).
Since P lies on the x-axis, its y-coordinate is 0.
Step 1: Use the condition of equidistance
Point P is equidistant from A and B.
So,
PA = PB
Step 2: Square the distances (distance formula without square root)
PA2 = (x − 7)2 + (0 − 6)2
PB2 = (x − 3)2 + (0 − 4)2
Step 3: Equate PA2 and PB2
(x − 7)2 + 36 = (x − 3)2 + 16
Step 4: Simplify
x2 − 14x + 49 + 36 = x2 − 6x + 9 + 16
x2 − 14x + 85 = x2 − 6x + 25
−14x + 85 = −6x + 25
−8x = −60
x = 7.5
Final Answer:
The point on the x-axis equidistant from (7, 6) and (3, 4) is
(7.5, 0)