The co-ordinates of the foot of the perpendicular from the point $(1,2)$ on the line $x-3 y+7=0$ are

The co-ordinates of the foot of the perpendicular from the point $(1,2)$ on the line $x-3 y+7=0$ are
  1. $\left(\frac{4}{5}, \frac{13}{5}\right)$
  2. $(-13,-2)$
  3. $\left(\frac{-13}{5}, \frac{-2}{5}\right)$
  4. $(2,3)$

Solution

$\begin{aligned} & \frac{x-1}{1}=\frac{y-2}{-3}=\frac{-(1-3 \times 2+7)}{1^2+(-3)^2} \\ & \Rightarrow \frac{x-1}{1}=\frac{y-2}{-3}=\frac{-2}{10} \\ & \Rightarrow x=\frac{4}{5} \text { and } y=\frac{13}{5} \\ & \Rightarrow\left(\frac{4}{5}, \frac{13}{5}\right)\end{aligned}$

Asked in: MHT CET 2022 (06 Aug Shift 1)

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