Given that the point P(k –1, 2) is equidistant from the points A(3, k) and B(k, 5).
Hence, distance of point P(k –1, 2) from point A(3, k)
= distance of point P(k –1, 2) from point B(k, 5).
= \(\sqrt{(3-k+1)^2+(k-2)^2}\) = \(\sqrt{(k-k+1)^2 + 3^2}\) (By distance formula)
⇒ (3 − k + 1)2 + (k − 2)2 = (k − k + 1)2 + 32 (By squaring both sides)
⇒ (4 − k)2 + (k − 2)2 = 1 + 9
⇒ 16 + k2 – 8k + k2 + 4 – 4k = 10 (∵(a − b)2 = a2 + b2 − 2ab)
⇒ 2k2 – 12k + 10 = 0
⇒ k2 – 6k + 5 = 0
⇒ (k – 5) (k – 1) = 0
⇒ k = 1 or k =5.
Hence, the values of k are k = 1 and k = 5
