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  4. If one diagonal of a square is along the line 8x - 15y = 0 and one of its vertex is at (1,2), then find the equation of the side of the square passing through this vertex.

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Q. If one diagonal of a square is along the line $8x - 15y = 0$ and one of its vertex is at $(1,2)$, then find the equation of the side of the square passing through this vertex.

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Solution:

Let $ABCD$ be the given square and the coordinates of the vertex $D$ be $(1, 2)$. We are required to find the equations of its sides $DC$ and $AD$.
Given that $BD$ is along the line $8x - 15y = 0$, so its slope is $\frac{8}{15}$. The angles made by $BD$ with sides $AD$ and $DC$ is $45^{\circ}$.

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Let the slope $DC$ be $m$. Then $tan\,45^{\circ}=\frac{m-\frac{8}{15}}{1+\frac{8m}{15}}$
or $15+8m=15m-8$
or $7m = 23$, which gives $m=\frac{23}{7}$
Therefore, the equation of the side $DC$ is given by
$y-2=\frac{23}{7}\left(x-1\right)$ or $23x-7y-9=0$.
Similarly, the equation of another side $AD$ is given by
$y-2=\frac{-7}{23}\left(x-1\right)$ or $7x+23y-53=0$