<s>[INST] Find the foot of the perpendicular through the point $A(-3;2)$ to the line $2x-y+4=0$. Find the foot of the perpendicular through the point $A(-3;2)$ to the line $2x-y+4=0$.
Let $AH\perp a:2x-y+4=0, H\in a.$
I have tried to use the fact that the vector $\vec{AH}(x_H+3;y_H-2)$ is a normal vector of the line $a$ and also $a$ passes through $H(x_H;y_H)$. Then we will have another equation for $a$ $$(x_H+3)(x-x_H)+(y_H-2)(y-y_H)=0$$
I am not familiar with the concept for slopes.
 [/INST] We have $y=2x+4$. Consider the line through $A$ 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 passes through $A$, $2=\frac{3}{2}+b\implies b=\frac{1}{2}$. 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:
$$2x+4=\frac{1}{2}-\frac{1}{2}x\implies 4x+8=1-x\implies x=-\frac{7}{5}$$
And then $y=-\frac{14}{5}+4=\frac{6}{5}$. So the answer is $\left(-\frac{7}{5}, \frac{6}{5}\right)$.
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