One side of a rectangle lies along the line 4x + 7y + 5 = 0. Two of its
vertices are (–3, 1) and (1,1). Find the equation of other three sides.

The point (-3, 1) lies on the line 4x + 7y + 5 = 0. So, let A(-3, 1) and C(1, 1) be 2 vertices of the rectangle ABCD.

Equation of line AB is 4x + 7y + 5 = 0.

CD is parallel to AB, passing through C(1, 1).

Let the equation of CD be 4x + 7y + k = 0........(1)

This passes through C(1, 1).

∴4(1) + 7(1) + k = 0

⇒k = -11

Putting the value of k in equation (1).

∴Equation of line CD is 4x + 7y -11 = 0.

As AD ⊥ AB, ∴equation of line AD is 7x - 4y + m = 0.......(2)

This passes through A(-3, 1).

∴7(-3) -4(1) + m = 0

⇒m = 25

Putting the value of m in equation (2).

∴Equation of line AD is 7x - 4y + 25 = 0.

As BC ⊥ AB, ∴equation of line BC is 7x - 4y + n = 0.......(3)

This passes through C(1, 1).

∴7(1) -4(1) + n = 0

⇒n = -3.

Putting the value of n in equation (3).

∴Equation of line BC is 7x - 4y - 3 = 0.