The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices
Let ABCD be a square having ( - 1, 2) and (3, 2) as vertices A and C respectively. Let (x, y), (x1, y1) be the coordinate of vertex B and D respectively
We know that the sides of a square are equal to each other
AB = BC
[(x + 1)2 + (y - 2)2]1/2 = [(x - 3)2 + (y - 2)2]1/2
x2 + 2x + 1 + y2 – 4y + 4 = x2 + 9 – 6x + y2 + 4 – 4y
8x = 8
x = 1
We know that in a square, all interior angles are of 90°.
In ΔABC,
AB2 + BC2 = AC2
[(x + 1)2 + (y - 2)2]1/2 * 2 + [(x - 3)2 + (y - 2)2]1/2 * 2 = [(3 + 1)2 + (2 - 2)2]1/2 * 2
4 + y2 + 4 - 4y + 4 + y2 - 4y + 4 =16
2y2 + 16 - 8 y =16
2y2 - 8 y = 0
y(y - 4) = 0
y = 0 or 4
We know that in a square, the diagonals are equal inlength and bisect each other at 90°.
Let O be the mid-point ofAC. Therefore, it will also be the mid-point of BD