Question

two opposite vertices of square r (-1,2)&(3,2) find coordinates of other two vertices

Answer 1

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 (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 ΔABC, we have

AB2 + BC2 = AC2 (Pythagoras theorem)

⇒ 2AB2 = AC2 (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 (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).