# Find the foot of the perpendicular through the point

Find the foot of the perpendicular through the point $A\left(-3;2\right)$ to line $2x-y+4=0$
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Athena Blanchard
Step 1
Any point on the line $2x-y+4=0$ is of the form $\left(t\in \mathbb{R}\right)$
Then slope of the line $AN:{m}_{1}=\frac{\left(2t+4\right)-2}{t-\left(-3\right)}=\frac{2t+2}{t+3}$
Now slope of the line $2x-y+4=0:{m}_{2}=2$
Two lines are perpendicular, hence ${m}_{1}\cdot {m}_{2}=-1$
$\frac{2t+2}{t+3}\cdot 2=-1$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}t=\frac{-7}{5}$
Hence, $N=\left(\frac{-7}{5},\frac{6}{5}\right)$
###### Not exactly what you’re looking for?
Step 1
We have $y=2x+4$ . Consider the line throudh A that is perpendicular to this line.
This line hence must have slope $-\frac{1}{2}$ .Thus its represented by $y=-\frac{1}{2}x+b$ for some b.
But since it passesthrough A, $2=\frac{3}{2}+b\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}b=\frac{1}{2}$ .
Thus the line is represented by $y=-\frac{1}{2}x+\frac{1}{2}$ . The foot of the perpendicular is simply the intersection of the two lines. Setting the y-coordinates, equal:
$2x+4=\frac{1}{2}-\frac{1}{2}x\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}4x+8=1-x\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=-\frac{7}{5}$
And then $y=-\frac{14}{5}+4=\frac{6}{5}$ . So the answer is $\left(-\frac{7}{5},\frac{6}{5}\right)$ .