If A(5, 2), B(2, – 2) and C(– 2, t) are the vertices of a right triangle with ∠B = 90°, then find the value of t.

From the fig we have B = 90°, so by Pythagoras theorem we have AC2 = AB2 + BC2

AC2 = (– 2 – 5)2 + (t – 2)2


= (– 7)2 + t2 + 4 – 2t


= 53 + t2 – 2t


AB2 = (2 – 5)2 + (– 2 – 2)2


= 9 + 16


= 25


BC2 = (– 2 – 2)2 + (t + 2)2


= 16 + t2 + 4 + 2t


= 20 + t2 + 2t


AB2 + BC2 = 45 + t2 + 2t


AC2 = 53 + t2 – 2t


53 + t2 – 2t = 45 + t2 + 2t


8 = 2t


t = 4


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