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Question

A = (7, -2) and C = (-1, -6) are the vertices of square ABCD. Find the equations of diagonals AC and BD.

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Solution

We know that in a square, diagonals bisect each other at right angle.

Let O be the point of intersection of the diagonals AC and BD.

Co-ordinates of O are

open parentheses fraction numerator 7 minus 1 over denominator 2 end fraction comma fraction numerator negative 2 minus 6 over denominator 2 end fraction close parentheses equals open parentheses 3 comma negative 4 close parentheses

Slope of AC = fraction numerator negative 6 plus 2 over denominator negative 1 minus 7 end fraction equals 1 half

For line AC:

Slope = m = 1 half, (x1, y1) = (7, -2)

Equation of the line AC is

y - y1 = m(x - x1)

y + 2 = 1 half (x - 7)

2y + 4 = x - 7

2y = x - 11

For line BD:

Slope = m = fraction numerator negative 1 over denominator s l o p e space o f space A C end fraction equals fraction numerator negative 2 over denominator blank end fraction, (x1, y1) = (3, -4)

Equation of the line BD is

y - y1 = m(x - x1)

y + 4 = -2(x - 3)

y + 4 = -2x + 6

2x + y = 2


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