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

Question

If a diagonal of a square is along the line $8 x-15 y=0$ and one of its vertices is $(1,2)$, then the equations of the sides of the square passing through this vertex are, $23 x-7 y+9=0,7 x+23 y+53=0$, $23 x-7 y-9=0,7 x+23 y-53=0$, $23 x+7 y-9=0,7 x+23 y-53=0$, $23 x+7 y-9=0,7 x+23 y+53=0$

MARKS App · Accepted Answer

Equation of line passes through point $(1,2)$ and inclined with an angle $45^{\circ}$ with respect to line $8 x-15 y=0$, because sides of a square inclined with diagonal at $45^{\circ}$, is $\begin{array}{rlrl} & y-2 & =\frac{8 / 15 \pm 1}{1 \mp\left(\frac{8}{15} \times 1\right)}(x-1) \ \Rightarrow & y-2 & =\frac{8 \pm 15}{15 \mp 8}(x-1) \ \Rightarrow \quad & y-2 & =\frac{23}{7}(x-1) \text { and } y-2=\frac{-7}{23}(x-1) \ \Rightarrow 23 x-7 y-9 & =0 \text { and } 7 x+23 y-53=0 \end{array}$

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