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.

Any point on the line $2x-y+4=0$ is of the the form $N=(t, 2t+4) $ $(t\in \Bbb{R}) $

Then slope of the line $AN : m_1=\frac{(2t+4)-2}{t-(-3) }=\frac{2t+2}{t+3} $

Now slope of the line $2x-y+4=0:m_2=2$

Two lines are perpendicular, hence $m_1 \cdot m_2=-1$

$\frac{2t+2}{t+3}\cdot 2=-1$

$\implies t=\frac{-7}{5}$

Hence, $N=(\frac{-7}{5},\frac{6}{5})$