The distance of a point $(2,5)$ from the line $3 x+y+4=0$ measured along the line $\mathrm{L}_1$ and…
- $\frac {-3}{4}$
- $\frac {1}{3}$
- $\frac {1}{4}$
- $0$
Solution
According to the given condition, in $\triangle \mathrm{ABC}$, $\mathrm{AB}=\mathrm{AC}$.
$\therefore \quad \triangle \mathrm{ABC}$ is an isosceles triangle.
Let $\mathrm{m}, \mathrm{m}_1, \mathrm{~m}_2$ be the slopes of given line, $\mathrm{L}_1$ and $\mathrm{L}_2$ respectively.
$\begin{array}{ll}
\therefore & \mathrm{m}=-3, \mathrm{~m}_1=\frac{3}{4} \\
\therefore & \left|\frac{\mathrm{m}-\mathrm{m}_1}{1+\mathrm{mm}_1}\right|=\left|\frac{\mathrm{m}-\mathrm{m}_2}{1+\mathrm{mm}_2}\right| \\
\therefore & 3=\left|\frac{-3-\mathrm{m}_2}{1+3 \mathrm{~m}_2}\right| \\
& \Rightarrow \mathrm{m}_2=0
\end{array}$Asked in: MHT CET 2023 (09 May Shift 1)