Question

P(−1, 2), A(2, k) and B(k, −1) are given points. If PA = PB, find the value of k.

Answer 1

Given:

Points P(−1, 2), A(2, k), and B(k, −1).

PA = PB; the distances from (P) to (A) and (P) to (B) are equal.

`"Distance" = sqrt((x_2 − x_1)^2 + (y_2 − y_1)^2)`

\[PA = \sqrt{(2 - (-1))^2 + (k - 2)^2}\] 

= \[\sqrt{(3)^2 + (k - 2)^2}\] 

= \[\sqrt{9 + (k - 2)^2} \]

= \[PB = \sqrt{(k - (-1))^2 + (-1 - 2)^2}\]

= \[\sqrt{(k + 1)^2 + (-3)^2}\] 

= \[\sqrt{(k + 1)^2 + 9}\]

Set (PA = PB):

= \[\sqrt{9 + (k - 2)^2}\] 

= \[\sqrt{(k + 1)^2 + 9}\]

Square both sides to eliminate the square roots

9 + (k − 2)² = (k + 1)² + 9

k − 2² = k + 1² ... [Simplify]

(k − 2)² = k² − 4k + 4

k − 2² = k² − 4k + 4 

\[ (k + 1)^2 = k^2 + 2k + 1\]     ...[Expand both sides]

= k² − 4k + 4 = k² + 2k + 1

= −4k + 4 = 2k + 1

= −4k − 2k = 1 − 4

= −6k = −3

\[k = \frac{-3}{-6} = \frac{1}{2}\]

Therefore, the value of k is `1/2`.