Find the foot of the perpendicular through the point $A(-3;2)$ to the line $2x-y+4=0$.

Question

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.

Answer 1

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

Answer 2

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 $\left(-\frac{7}{5}, \frac{6}{5}\right)$.