Question

One side of a rectangle lies along the line 3x + 7y + 2 = 0 and (−3, 1) is a vertex on it. If (1, 1) is another vertex of the rectangle, find the equations of the other three sides.

Answer 1

Let ABCD be the rectangle (see Fig.) where A is (−3, 1) and the equation of AD is 3x + 7y + 2 = 0. Since the line joining (−3, 1) and (1, 1) is horizontal, the point (1, 1) must be the end of the diagonal through A (−3, 1). Therefore, C is (1, 1) and the equation of BC is

y - 1 = -3/7(x - 1)

3x + 7y - 10 = 0

Equation of AB is

y - 1 = 7/3(x + 3)

⇒ 7x  - 3y  24 = 0

Equation of the side CD is

y - 1 = 7/3(x - 1)  ⇒ 7x  - 3y - 4 = 0

Therefore, the other three sides are 3x + 7y − 10 = 0, 7x − 3y + 24 = 0 and 7x − 3y − 4 = 0.