The co-ordinates of the foot of perpendicular, drawn from the point $(-2,3)$ on the line $3 x-y-1=0$ are
- $(-1,2)$
- $(1,-2)$
- $(-1,-2)$
- $(1,2)$
Solution
Slope of $3 x-y-1=0$ is 3 . Line passing through $(\mathrm{h}, \mathrm{k})$ and $(-2,3)$ is perpendicular to $3 x-y-1=0$. $\begin{aligned} & \therefore \quad \frac{k-3}{h+2} \times 3=-1 \\ & \Rightarrow \mathrm{~h}+3 \mathrm{k}=7...(ii) \end{aligned}$
Solving (i) and (ii), we get $\mathrm{h}=1$ and $\mathrm{k}=2$
Asked in: MHT CET 2024 (16 May Shift 2)