1.

If the point P (2, 2) is equidistant from the points A (- 2, k) and B (- 2k, -3), find k. Also, find the length of AP.

Answer»

Coordinates of points are P(2, 2) A(- 2, k) and B(- 2k, -3) 

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

⇒ PA = PB

\(\sqrt{(-2 - 2)^2 + (k - 2)^2}\) = \(\sqrt{(-2k - 2)^2 + (-3 - 2)^2}\)

On squaring both sides, we get

(- 2 - 2)2 + (k - 2)2 = (- 2k - 2)2 + (- 3 - 2)2

16 + k2 - 4k + 4 = 4k2 + 8k + 4 + 25

k2 + 4k + 3 = 0

k2 + 3k + k + 3 = 0

k(k + 3) + 1(k + 3) = 0

(k + 1)(k + 3) = 0

k = -1,- 3

ap = \(\sqrt{(-2 - 2)^2 + (-1 - 2)^2}\) = \(\sqrt{16 + 9}\) = 5 units



Discussion

No Comment Found

Related InterviewSolutions