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