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…
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
\(23 x-7 y+9=0,7 x+23 y+53=0\)
\(23 x-7 y-9=0,7 x+23 y-53=0\)
\(23 x+7 y-9=0,7 x+23 y-53=0\)
\(23 x+7 y-9=0,7 x+23 y+53=0\)
Solution
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
\(\begin{array}{rlrl}
& y-2 & =\frac{8 / 15 \pm 1}{1 \mp\left(\frac{8}{15} \times 1\right)}(x-1) \\
\Rightarrow & y-2 & =\frac{8 \pm 15}{15 \mp 8}(x-1) \\
\Rightarrow \quad & y-2 & =\frac{23}{7}(x-1) \text { and } y-2=\frac{-7}{23}(x-1) \\
\Rightarrow 23 x-7 y-9 & =0 \text { and } 7 x+23 y-53=0
\end{array}\)