Find the foot of the perpendicular through the point

Step 1
Any point on the line $2x-y+4=0$ is of the form $(t\in \mathbb{R})$
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 $m_1\cdot m_2=-1$
$\frac{2t+2}{t+3}\cdot 2=-1$
$⟹t=\frac{-7}{5}$
Hence, $N=(\frac{-7}{5},\frac{6}{5})$

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, $2=\frac{3}{2}+b⟹b=\frac{1}{2}$ .
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:
$2x+4=\frac{1}{2}-\frac{1}{2}x⟹4x+8=1-x⟹x=-\frac{7}{5}$
And then $y=-\frac{14}{5}+4=\frac{6}{5}$ . So the answer is $(-\frac{7}{5},\frac{6}{5})$ .