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The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices. (Ans: (1, 0), (1, 4))
Ans : AB = BC ⇒ AD2 = BC2
(x + 1)2 + (y-2)2 = (x-3)2 + (y-2)2
x2 + 2x + 1 + y2 - 4y + 4 = x2 - 6x + 9 + y2 - 4y + 4
2x - 4y + 5 = - 6x - 4y + 13
8x = 13 - 5 8x = 8 ⇒ x = 1
On substituting in (x-3)2 + (y-2)2 + (x+1)2 + (y-2)2
= (-1 -3)2 + (2 - 2)2
We get y = 4 or 0.
∴B (1, 4) or (1, 0)
AD = DC ⇒ AD2 = DC2
(x1 + 1)2 + (y1-2)2 = (x1-3)2 + (y1-2)2
∴ x = 1.
On substituting in (x1 + 1)2 + (y1-2)2 + (x1-3)2 + (y1-2)2 = 16
We get y1 = 0 or 4.
∴ D (1, 4) or (1, 0)
∴the opposite vertices are (1, 4) & (1, 0)
simplex methord
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