1.

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

Answer»

The coordinates are A(-1, 2) and C(3, 2). 

Let the coordinates of the vertex B are (x, y) 

AB = BC 

Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

\(\sqrt{(x + 1)^2 + (y - 2)^2}\) = \(\sqrt{(x - 3)^2 + (y - 2)^2}\)

On squaring both sides, we get

(x + 1)2 + (y - 2)2 = (x - 3)2 + (y - 2)2

x2 + 2x + 1 + y2 - 4y + 4 = x2 - 6x + y2 - 4y + 4

x = 1

In ΔABC 

AB2 + BC2 = AC2 [Using Pythagoras theorem] 

2AB2 = AC2 [Since AB = BC]

2[(x + 1)2 + (y - 2)2] = (3 + 1)2 + (2 - 2)2

2[x2 + 2x + 1 + y2 - 4y + 4] = 16

x2 + 2x + 1 + y2 - 4y + 4 = 8

x2 + 2x + y2 - 4y = 3

On substituting x = 1

1 + 2 x 1 + y2 - 4y = 3

y2 - 4y = 0

y(y - 4) = 0

y = 0, 4

Other coordinates are (1, 0) and (1, 4)


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