One side of a rectangle lies along the line `4x+7y+5=0.` Two of its vertices are `(-3,1)a n d(1,1)dot` Find the equations of the other three sides.

Explanation

Let the side AB of rectangle ABCD lies along 4x+7y+5 = 0. As (-3,1) lies on this line, let it be vertex A. Now, (1,1) is either vertex C or D.

If (1,1) is vertex D, then the slope of AD is 0. Hence, AD is not perpendicular to AB. But it is a contradiction as ABCD is a rectangle. Therefore, (1,1) are the coordinates of vertex C.
CD is a line parallel to AB and passing through C. Therefore, the equation of CD is
`y-1 = -(4)/(7)(x-1)`
or 4x+7y-11=0
Also, BC is a line perpendicular to AB and passing through C.
Therefore, the equation of BC is
`y-1 = (7)/(4)(x-1)`
or 7x-4y-3=0
Similarly, AD is a line perpendicular to AB and passing through A(-3,1). Therefore, the equation of line AD is
`y-1 = (7)/(4)(x+3)`
or 7x-4y+25 = 0