Question

If the point P(k - 1 , 2 ) is equidistant from the points A (3, k) and b (k, 5), find the value of k​

Answer 1

Step-by-step explanation:

The point P(k - 1 , 2 ) is equidistant from the points A (3, k) and B(k, 5)

Distance of PA = Distance of PB

$distance = \sqrt{ {(x2 - x1)}^{2} + (y2 - y1)^{2} }$

$\sqrt{ {(k - 1 - 3)}^{2} + {(2 - k)}^{2} } = \sqrt{ {(k - 1 - k)}^{2} + {(2 - 5)}^{2} }$

On squaring on both sides, square roots would be cancelled.

$(k - 4)^{2} + (2 - k)^{2} = 1 + 9$

${k}^{2} - 8k + 16 + {k}^{2} + 4 - 4k = 10$

$2k^{2} - 12k + 10 = 0$

$(k - 5)(k - 1) = 0$

$k = 5;k = 1$