Exercise 324Geometry › Analytic geometryHigh schoolAdded 31 Oct 2026

Distance from a point to a line

#lines#projection

Statement

Compute the distance from the point $M(1,2)$ to the line $\mathcal D:\ 3x+4y-5=0$ .

Hint

The vector $\vec n=(3,4)$ is normal to $\mathcal D$ . Project $M$ onto $\mathcal D$ along $\vec n$ .

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Solution

The foot $H$ of the perpendicular from $M$ has the form $H=M-t\,\vec n$ . Requiring $H\in\mathcal D$ :
$3(1-3t)+4(2-4t)-5=0\quad\Longrightarrow\quad6-25t=0\quad\Longrightarrow\quad t=\frac6{25}.$

Then $MH=|t|\,\|\vec n\|=\dfrac6{25}\cdot5=\dfrac65$ .

This agrees with the general formula $d=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}=\dfrac{|3+8-5|}{5}=\dfrac65$ .

Distance from a point to a line

Statement

Solution

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