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.**

**
**