Why is vector $[A,B]$ perpendicular to line $Ax + By + C = 0$?

Question

Why is vector $[A,B]$ perpendicular to line $Ax + By + C = 0$ and vector $[-B, A]$ is parallel to it? I've seen some proofs, but I don't know what is the intuition behind this. After thinking about this form of representation of a line I can't see why this equation represents a line and why does those vectors are perpendicular/parallel. Can someone give me some intuition how to think about it?

Answer 1

Write it as,

$$A(x-h)+B(y-k)=0$$

Let $\vec v=\langle A,B \rangle$ and $\vec w= \langle h,k \rangle$ and $\vec x=\langle x,y \rangle$.

Then,

$$\vec v \cdot (\vec x-\vec w)=0$$

The tail of position vector $\vec w$ lies on the line $Ax+By=C$. If you draw $\vec x-\vec w$ you will see why this shows $\vec v$ is orthogonal to the line.

Now as two why $\langle -B, A \rangle$ is a direction vector. You can deduce this from the fact that $\langle -B,A \rangle \cdot \langle A,B \rangle=0$.

Answer 2

For example take two points in the line, $P=(0,-C/B)$ and $Q=(-C/A,0)$ so the direction vector can be:

$\vec{PQ}=Q-P=(-C/A,C/B)$

then

$$(A,B)\cdot(-C/A,C/B)=-C+C=0\to(A,B)\perp \vec{PQ}$$

$$(-B,A)= \frac{A\cdot B}{C}\cdot \left(-\frac{C}{A},\frac{C}{B}\right)\to (-B,A)||(\vec{PQ})$$

Answer 3

By definition $[A,B]=B\vec{j}-A\vec{i}$, also the vector of line is $\vec{v}=A\vec{i}+B\vec{j}$ thus $$[A,B].\vec{v}=(B\vec{j}-A\vec{i}).(A\vec{i}+B\vec{j})=BA-AB=0$$ so they are perpendicular.

Answer 4

One way to write an equation of a line is $\mathbf n\cdot\mathbf x=d$. That is, the line consists of all points for which the dot product of their position vector with some fixed vector $\mathbf n$ is constant. Now, this dot product is just the orthogonal projection of $\mathbf x$ onto $\mathbf n$, scaled by the length of $\mathbf n$, so another way to put this is that the position vector of every point on the oine has the same projection onto $\mathbf n$. This means that the position vector of every point on the line can be decomposed into $\mathbf n_\parallel+t\mathbf n_\perp$, where $\mathbf n_\parallel$ is some fixed multiple of $\mathbf n$ that’s on the line (it’s not hard to work out exactly what $\mathbf n_\parallel$ is) and $\mathbf n_\perp$ is some fixed vector perpendicular to $\mathbf n$. The latter vector is a clearly a direction vector for the line, so we can conclude that $\mathbf n$ is perpendicular to the line. (By the way, the length of $\mathbf n_\parallel$ is the distance of the line from the origin. Can you see why?)

Now, rewrite the equation $Ax+By+C=0$ in the above form: $[A,B]\cdot[x,y]=-C$. From the above discussion, we see that the vector $\mathbf n=[A,B]$ is perpendicular to the line. Since $[A,B]\cdot[-B,A]=0$, the vector $[-B,A]$ is perpendicular to $\mathbf n$ and so is parallel to the line. Any vector parallel to the line works as a direction vector for it, so $[-B,A]$ is a direction vector.

You can also find directly that $[-B,A]$ is a direction vector. By definition, a direction vector for the line is parallel to it, so in particular they have the same slope. The slope of the line is easily found to be $-A/B$ (ignoring for the moment the uncomfortable case $B=0$) so we can read off the components of a parallel vector from this fraction: $[-B,A]$. As before, $[A,B]\cdot[-B,A]=0$, so $[A,B]$ is perpendicular to $[-B,A]$ and thus perpendicular to the line as well.