Question:

The foot of the perpendicular from $(-2,3)$ to the line $2x - y - 3 = 0$ is

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Foot of perpendicular from $(x_1,y_1)$ to $ax+by+c=0$ is $(x_1-a\frac{ax_1+by_1+c}{a^2+b^2}, y_1-b\frac{ax_1+by_1+c}{a^2+b^2})$.
Updated On: Apr 8, 2026
  • $(-2,3)$
  • $(2,1)$
  • $(3,2)$
  • $(1,2)$
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The Correct Option is B

Solution and Explanation

Step 1: Foot formula: $(x_1,y_1) = (-2,3)$, line $2x-y-3=0$.}
Step 2: $\frac{x+2}{2} = \frac{y-3}{-1} = -\frac{2(-2)-3-3}{5} = -\frac{-4-6}{5} = \frac{10}{5}=2$. So $x=2$, $y=1$.}
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