Answer
See the explanation below.
Work Step by Step
Express the line equation passing through these 2 distinct points in the form of matrix “A” as follows:
$A=\left[ \begin{matrix}
x & y & 1 \\
{{x}_{1}} & {{y}_{1}} & 1 \\
{{x}_{2}} & {{y}_{2}} & 1 \\
\end{matrix} \right]$
To obtain the equation of the line, put $Det\left( A \right)=0$:
$\begin{align}
& x\left\{ \left( {{y}_{1}}.1 \right)-\left( {{y}_{2}}.1 \right) \right\}-y\left\{ \left( {{x}_{1}}.1 \right)-\left( {{x}_{2}}.1 \right) \right\}+1\left\{ \left( {{x}_{1}}{{y}_{2}} \right)-\left( {{x}_{2}}{{y}_{1}} \right) \right\}=0 \\
& x\left( {{y}_{1}}-{{y}_{2}} \right)-y\left( {{x}_{1}}-{{x}_{2}} \right)+\left( {{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}} \right)=0 \\
& x\left( {{y}_{1}}-{{y}_{2}} \right)+\left( {{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}} \right)=y\left( {{x}_{1}}-{{x}_{2}} \right) \\
& y=\left( \frac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}} \right)x+\left( \frac{{{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}}}{{{x}_{1}}-{{x}_{2}}} \right)
\end{align}$
Therefore, the equation now has obtained a form of $y=mx+c$, where
$m=\left( \frac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}} \right)\text{ and }c=\left( \frac{{{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}}}{{{x}_{1}}-{{x}_{2}}} \right)$
Hence, the determinant equation can be used to give the equation if the line is passing through the points $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$.
