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Prove that a vector perpendicular to the line $Ax+By+C=0$ in the $xy$-plane is $<A, B>$

So intuitively this is clear, since we can always find a vector normal to a line by looking at the coefficients of $x,y,z$ So in this case $<A, B, 0>$ or $<A, B>$ in 2d space is clearly a normal vector of the line.

I tried using this approach but got stuck : take two points on the line, find a vector parallel to the line, take the dot product with $<A, B>$ and show this is $= 0$
So
Let $P_1 = Ax_1 + By_1 = -C$ be a point on the line and
Let $P_2 = Ax_2 + By_2 = -C$ be another point on the line.
(let this denote the vector) Then $P_1P_2 = <Ax_2 - Ax_1, By_2 - By_1>$ is parallel to the line.
So

$<A, B> \cdot P_1P_2 = A^2x_2-A^2x_1+B^2y_n-B^2y_1$
$= A^2x_2 + B^2y_n - (A^2x_1+B^2y_1)$
But I wasn't able to get this to equal 0 using the fact that $Ax + By = -C$
Any help is appreciated!
Thanks.

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2 Answers 2

up vote 2 down vote accepted

The only problem you have is that your expression for the vector $\overrightarrow{P_1P_2}$ is slightly off. As you say, let $P_1 = (x_1,y_1)$ and $P_2 = (x_2,y_2)$ be points on the line. Then what are the coordinates of the vector $\overrightarrow{P_1P_2}$?

Once you calculate that, then notice that the coordinates of $P_1$ satisfy the equation $Ax_1 + By_1 = -C$, and similarly for $P_2$. Then calculate the scalar product of $(A,B)$ with the vector $\overrightarrow{P_1P_2}$, and interpret what comes out.

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Thanks, it worked out! –  fdart17 Dec 5 '10 at 23:00

vector is equals $<x_2 - x_1, y_2 - y_1>$, not $<Ax_2 - Ax_1, By_2 - By_1>$ then $<A,B> \cdot P_1P_2 = A\cdot x_2-A\cdot x_1+B\cdot y_2-B\cdot y_1$

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