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 A (1, 2) be the vertex of square ABCD and BD be the diagonal that is along the line 8x − 15y = 0

$$\begin{gathered} \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{line}\mathrm{is}, \\ 8x-15y=0 \\ \Rightarrow -15y=-8x \\ \Rightarrow y=\frac{8}{15}x \\ \mathrm{Comparing}\mathrm{this}\mathrm{equation}\mathrm{with}\mathit{}y=mx+c \\ \mathrm{we}\mathrm{get}, \\ m=\frac{8}{15} \end{gathered}$$
So, the slope of BD will be $\frac{8}{15}$.


Here, we have to find the equations of sides AB and AD.

We know that the equations of two lines passing through a point $\left(x_1,y_1\right)$ and making an angle $\mathrm{\alpha }$ with the line whose slope is m.

$y-y_1=\frac{m\pm \tan \alpha }{1\mp m\tan \alpha }\left(x-x_1\right)$

Here,

$$\begin{gathered} m=\frac{8}{15} \\ {x}_1=1 \\ {y}_1=2 \\ \mathrm{\alpha }={45}^{\circ } \end{gathered}$$

So, the equations of the required sides are

$$\begin{gathered} y-2=\frac{{\displaystyle \frac{{\displaystyle 8}}{15}+\tan {45}^{\circ }}}{{\displaystyle 1-\frac{{\displaystyle 8}}{15}\tan {45}^{\circ }}}\left(x-1\right)\mathrm{and}y-2=\frac{{\displaystyle \frac{{\displaystyle 8}}{15}-\tan {45}^{\circ }}}{{\displaystyle 1+\frac{{\displaystyle 8}}{15}\tan {45}^{\circ }}}\left(x-1\right) \\ \Rightarrow y-2=\frac{{\displaystyle \frac{{\displaystyle 8}}{15}+1}}{{\displaystyle 1-\frac{{\displaystyle 8}}{15}}}\left(x-1\right)\mathrm{and}y-2=\frac{{\displaystyle \frac{{\displaystyle 8}}{15}-1}}{{\displaystyle 1+\frac{{\displaystyle 8}}{15}}}\left(x-1\right) \\ \Rightarrow 23x-7y-9=0\mathrm{and}7x+23y-53=0 \end{gathered}$$