Question

If the p(K-1,2) IS equidistant from the point A(3,k) and B(k,5),find thw values of k

Answer 1

Given that PA = PB

Squaring both sides

PA² = PB²

[3-(k-1)]²+[k-2]² = [k-(k-1)]²+[5-2]²       [ using distance formula ]

⇒ 9+(k-1)²-6(k-1)+(k-2)² = k²+(k-1)²-2k(k-1)+9

⇒ 9-6(k-1)+(k-2)² = k²-2k(k-1)+9

⇒ 9-6k+6+k²+4-4k = k²-2k²+2k+9

⇒ 2k²-12k+10 = 0

⇒ k²-6k+5 = 0

⇒ k²-k-5k+5 = 0

⇒ k(k-1)+5(k-1) = 0

⇒ (k-1)(k+5) = 0

⇒ k = 1   or   k= -5