1.

If the point P (x, 3) is equidistant from the points A (7,-1) and B (6, 8), find the value of x and find the distance AP.

Answer»

Coordinates are A (7,-1) and B (6, 8) 

The point P (x, 3) is equidistant. 

Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

⇒ PA = PB

\(\sqrt{(x - 7)^2 + (3 + 1)^2}\) = \(\sqrt{(x - 6)^2 + (3 - 8)^2}\)

On squaring both sides, we get

(x - 7)2 + (3 + 1)2 = (x - 6)2 + (3 - 8)2

x2 - 14x + 49 + 16 = x2 - 12x + 36 + 25

x = 2

AP = \(\sqrt{(x - 7)^2 + (3 + 1)^2}\) = \(\sqrt{25 + 16}\) = \(\sqrt{41}\) units



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