Question

The diagonal of a square lies along the line $8x-15y=0$ and one vertex of the square is (1, 2). Find the equations to the sides of the square passing through this vertex.

Answer 1

Let BD be the diagonal $8x-15y=0$ of square ABCD

Let slope of AB be m and slope of BD is $\frac{8}{15}$ then,

$tan{45}^{\circ }=\frac{m-\frac{8}{15}}{1+\frac{8}{15}}m$

$15-8m=15m-8$

$7m=23$

$m=\frac{23}{7}$

and, (slope of BC) $\times$ (slope of AB) = - 1

slope of BC $=-1\times \frac{7}{23}=\frac{-7}{23}$

$\therefore$ Equation of AB is

$y-2=\frac{23}{7}(x-1)$

or, $23x-7y-9=0$

And, equation of BC is

$y-2=\frac{-7}{23}(x-1)$

or $7x+23y-53=0$