Two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of other two vertices.

Given that A(- 1, 2) and C(3, 2) are the opposite vertices of a square.



Let us assume the other two vertices be B(x1, y1) and D(x2, y2) and the midpoint be M


We know that midpoint of AC = Midpoint of BD = M




⇒ M = (1, 2)



⇒ x1 + x2 = 2 ..... (1)


⇒ y1 + y2 = 4 ..... (2)


We know that lengths of the sides of the square are equal.


AB = BC = CD = DA.


We know that distance between two points (x1, y1) and (x2, y2) is


⇒ AB = BC


⇒ AB2 = BC2


⇒ (x1 - (- 1))2 + (y1 - 2)2 = (x1 - 3)2 + (y1 - 2)2


⇒ x12 + 1 + 2x1 + y12 + 4 - 4y1 = x12 - 6x1 + 9 + y12 + 4 - 4y1


⇒ 8x1 = 8



⇒ x1 = 1 ..... (3)


From (1)


⇒ x2 = 2 - 1 = 1 ..... (4)


We know that points ABC form right angled isosceles triangle.


We have AB2 + BC2 = AC2


⇒ 2AB2 = (- 1 - 3)2 + (2 - 2)2


⇒ 2((1 - (- 1))2 + (y1 - 2)2) = (- 4)2 + (0)2


⇒ 2(22 + (y1 - 2)2) = 8


⇒ 4 + (y1 - 2)2 = 8


⇒ (y1 - 2)2 = 4


⇒ y1 - 2 = ±2


⇒ y1 = 2 - 2 (or) y1 = 2 + 2


⇒ y1 = 0 (or) y1 = 4


From (2)


⇒ y2 = 4 - 0


⇒ y2 = 4


⇒ y2 = 4 - 4


⇒ y2 = 0


It is clear that the other two points are (1, 0) and (1, 4).


∴ The other two points are (1, 0) and (1, 4).


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