Q90 / 150 · MHT CET 2024 (16 May Shift 2) · Mathematics
The co-ordinates of the foot of perpendicular, drawn from the point $(-2,3)$ on the line $3 x-y-1=0$ are
- A$(-1,2)$
- B$(1,-2)$
- C$(-1,-2)$
- Correct answerD$(1,2)$
Solution
Let $(h, k)$ be the required point on the line $3 x-y-1=0$.
$\therefore \quad 3 \mathrm{~h}-\mathrm{k}-1=0...(i)$
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 $\mathrm{k}=2$
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 $\mathrm{k}=2$
Hence, the correct answer is option: 'D'.