Answer
See proof
Work Step by Step
Consider the equation:
$\begin{vmatrix}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{vmatrix}=0$
Expand the determinant:
$x\begin{vmatrix}y_1&1\\y_2&1\end{vmatrix}-y\begin{vmatrix}x_1&1\\x_2&1\end{vmatrix}+1\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}=0$
$x(y_1-y_2)-y(x_1-x_2)+(x_1y_2-x_2y_1)=0$
The above equation describes a line. We will check if the points $(x_1,y_1)$ and $(x_2,y_2)$ belong to this line:
$(x_1,y_1)$;
$x_1(y_1-y_2)-y_1(x_1-x_2)+x_1y_2-x_2y_1=x_1y_1-x_1y_2-x_1y_1+x_2y_1+x_1y_2-x_2y_1=0\checkmark$
$(x_2,y_2)$;
$x_2(y_1-y_2)-y_2(x_1-x_2)+x_1y_2-x_2y_1=x_2y_1-x_2y_2-x_1y_2+x_2y_2+x_1y_2-x_2y_1=0\checkmark$
Therefore the two points belong to the line. Because a line is determined by two points belonging to it, it means that the equation describes the line passing through the two points.
