Question

One diagonal of a square is along the line $8x-15y=0$ and one of its vertices is $(1,2)$. Then the equations of the sides of the square passing through this vertex are

• A. $23x+7y=9,7x+23y=53$
• B. $23x-7y+9=0,7x+23y+53=0$
• C. $23x-7y-9=0,7x+23y-53=0$
• D. None of these

Answer 1

Answer: (C)

$23x-7y-9=0,7x+23y-53=0$

Let the square be $ABCD$ and vertex $C$ be $(1,2)$ . Slope of $BD$ is $\frac{8}{15}$ and angle made by $BD$ with $DC$ and $BC$ is $45$${}^{\circ }$. So let slope of $DC$ be $m$. Then,

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

$\Rightarrow (15+8m)=\pm (15m-8)$

$\Rightarrow m=\frac{23}{7}$ and $-\frac{7}{23}$

Hence, the equations of $DC$ and $BC$ are

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

$\Rightarrow 23x-7y-9=0$

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

$\Rightarrow 7x+23y-53=0$