Equations of the diagonals of a rectangle are y + 8x – 17 = 0 and y – 8x + 7 = 0. If the area of the rectangle is 8 sq. units, then the equation of the sides of the rectangle is/are

The intersection point of the given diagonals is $P\equiv \left(\frac{3}{2},5\right)$

Equation of angular bisectors of the diagonals are $\frac{y+8x-17}{\sqrt{65}}=\pm \frac{y-8x+7}{\sqrt{65}}$

$\Rightarrow x=\frac{3}{2} and y=5$

Let length of BC be a and that of CD be b          

Then tan q $=\frac{a/2}{b/2}=\frac{a}{b}=8$

Also ab = 8

Þ  a = 8, b = 1.

So equations of sides are y = 1, y = 9, x = 1 and x = 2.