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

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

So with ease this is clear, given that we can constantly locate a vector regular to a line by considering the coefficients of $x,y,z$ So in this instance $<A, B, 0>$ or $<A, B>$ in 2d room is plainly a regular vector of the line.

I attempted utilizing this strategy yet obtained stuck : take 2 factors on the line, locate a vector alongside the line, take the dot item with $<A, B>$ and also show this is $= 0$

So

Let $P_1 = Ax_1 + By_1 = -C$ be a factor on the line and also

Let $P_2 = Ax_2 + By_2 = -C$ be an additional factor on the line.

(allow this represent the vector) Then $P_1P_2 = <Ax_2 - Ax_1, By_2 - By_1>$ is alongside 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 had not been able to get this to equivalent 0 making use of the reality that $Ax + By = -C$

Any aid is valued!

Many thanks.

The only trouble you have is that your expression for the vector $\overrightarrow{P_1P_2}$ is a little off. As you claim, allow $P_1 = (x_1,y_1)$ and also $P_2 = (x_2,y_2)$ be factors on the line. After that what are the works with of the vector $\overrightarrow{P_1P_2}$?

As soon as you compute that, after that see that the works with of $P_1$ please the formula $Ax_1 + By_1 = -C$, and also in a similar way for $P_2$. After that compute the scalar item of $(A,B)$ with the vector $\overrightarrow{P_1P_2}$, and also analyze what appears.

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