Prove that if the line $\mathfrak{L}$ is the graph of the linear Cartesian equation $A x+B y+C=0, A^2+B^2 \neq 0$, and $\mathbf{v}$ is a vector having a geometric representation with one endpoint on $£$ and the other not on $\mathcal{L}$, then $\mathbf{v}$ is not parallel to any nonzero vector having a geometric representation with both endpoints on \&. [Hint: Let $\mathbf{S}\left(x_1, y_1\right)$ be any point on $£$ and $\mathbf{T}\left(x_2, y_2\right)$ be any point not on $£$. Show that $A\left(x_2-x_1\right)+B\left(y_2-y_1\right) \neq 0$, and accordingly that the vector $\left(x_2-x_1, y_2-y_1\right)$ is not perpendicular to the vector $(A, B)$.]