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Question
Find a point on x axis which is equidistant from the points (7, –6) and (3, 4).
Solution
Let P(x1, 0) be any point on the x-axis.0
Let A(7, -6) and B(3, 4) be the given points.
Given that PA = PB
PA2 = PB2
(x1 - 7)2 + (0 + 6)2 = (x1 - 3)2 + (0 - 4)2
`x_1^2 - 14x_1 + 49 + 36 = x_1^2 - 6x_1 + 9 + 16`
- 14x1 + 6x1 = 25 - 85
- 8x1 = - 60
`x_1 = (-60)/(-8) = 15/2`
∴ The required point is `(15/2, 0)`
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