Answer
See the description below.
Work Step by Step
Showing that the equation of the line in $\mathbb { R } ^ { 2 }$ through distinct points $(x_{1},y_{1})\,and\,(x_{2}, y_{2}) $
Given the equation as;
$det\begin{bmatrix}1&x&y\\1&x_{1}&y_{1}\\1&x_{2}&y_{2}\end{bmatrix}=0$
Finding the determinant along the first row we have;
$1(x_{1}y_{2}-x_{2}y_{1})-x(y_{2}-y_{1})+y(x_{2}-x_{1})$
$x_{1}y_{2}-x_{2}y_{1}-xy_{2}+xy_{1}+yx_{2}-yx_{1}=0$ ..............................................eqn 1
By using the equation of a straight line.
$\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Re-arranging the equation and expanding;
$(y-y_{1})(x_{2}-x_{1})=(y_{2}-y_{1})(x-x_{1})$
$yx_{2}-yx_{1}-y_{1}x_{2}+y_{1}x_{1}=x y_{2}-x_{1}y_{2} -xy_{1}+x_{1}y_{1}$
$yx_{2}-yx_{1}-y_{1}x_{2}-x y_{2}+x_{1}y_{2}+xy_{1}=0$.................................................eqn 2
eqn1 and eqn2 are equal. The equation of the line in $\mathbb { R } ^ { 2 }$ through distinct points $(x_{1},y_{1})\,and\,(x_{2}, y_{2}) $ can also be represented as;
$det\begin{bmatrix}1&x&y\\1&x_{1}&y_{1}\\1&x_{2}&y_{2}\end{bmatrix}=0$
