Answer
$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$
Work Step by Step
The determinant for a 3 by 3 matrix can be computed as: $Determinant (D)= \begin{vmatrix} a & b & c \\
d & e & f \\
g & h & i \\ \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)$
We have:
$D= \begin{vmatrix}
x & y & 1 \\
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\ \end{vmatrix}
$
Therefore, $D=x (y_1-y_2)-y(x_1-x_2)+(x_1y_2-x_2y_1) $
or, $xy_1-xy_2-yx_1+yx_2+x_1y_2-x_2y_1 =0$
or, $(y_2-y_1)(x-x_1)=(y-y_1)(x_2-x_1)$
or, $y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$
This represents two points form an equation of a straight line.
