CBSE Class 10 Answered
Let ABCD be a square and A (1, 2) and C (3, 2) be the given vertices. Let the coordinate of vertex B be (x,y).
AB = BC (as ABCD is a square)
=> AB2 = BC2
=> [x (1)] 2 + (y 2)2 = (x 3)2 + (y 2)2 (Distance formula)
=> (x + 1)2 = (x 3)2
=> x2 + 2x + 1 = x2 6x + 9
=> 2x + 6x = 9 1
=> 8x = 8
=> x = 1
In triangle ABC, we have
AB2 + BC2 = AC2 (Pythagoras theorem)
=> 2AB2 = AC2 (as AB = BC)
=> 2[(x (1))2 + (y 2)2] = (3 (1))2 + (2 2)2
=> 2[(x + 1)2 + (y 2)2] = (4)2 + (0)2
=> 2[(1 + 1)2 + (y 2)2] = 16 (as x = 1)
=> 2[ 4 + (y 2)2] = 16
=> 8 + 2 (y 2)2 = 16
=> 2 (y 2)2 = 16 8 = 8
=> (y 2)2 = 4
=> y 2 = ± 2
=> y 2 = 2 or y 2 = 2
=> y = 4 or y = 0
Thus, the other two vertices of the square ABCD are (1, 4) and (1, 0).