Answer
$[(x-x_0) i+(y-y_0) j] \cdot (Ai+Bj) =A(x-x_0) +B(y-y_0)=0$
The result has been proved.
Work Step by Step
The equation for a line vector is:
$v=(x-x_0) i+(y-y_0) j$
The line is passing through the points $(x_0,y_0)$ and $(x,y)$ and the vector $n = Ai+Bj$ is normal to it.
So $v \cdot N =0$
Thus, we find that
$[(x-x_0) i+(y-y_0) j] \cdot (Ai+Bj) =A(x-x_0) +B(y-y_0)=0$
The result has been proved.
