Question

Consider two distinct points$\left[\begin{array}{l}{a}_1\\ a_2\end{array}\right]$ and$\left[\begin{array}{l}{b}_1\\ b_2\end{array}\right]$ in the plane. Explain why the solutions$\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]$ of the equation$\mathit{d}\mathit{e}\mathit{t}\left[\begin{array}{ccc}1& 1& 1\\ x_1& a_1& b_1\\ x_2& a_2& b_2\end{array}\right]\mathbf{=}\mathbf{0}$ form a line and why this line goes through the two points$\left[\begin{array}{l}{a}_1\\ a_2\end{array}\right]$ and $\left[\begin{array}{l}{b}_1\\ b_2\end{array}\right]$.

Short answer

Therefore, $\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{l}{a}_1\\ a_2\end{array}\right]$ and $\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{l}{b}_1\\ b_2\end{array}\right]$ are a solution.

Step-by-step solution

Given

Consider two distinct points are,

$\left[\begin{array}{l}{a}_1\\ a_2\end{array}\right]$

$\left[\begin{array}{l}{b}_1\\ b_2\end{array}\right]$

To solve equations 

We solve,

$\det \left[\begin{array}{ccc}1& 1& 1\\ x_1& a_1& b_1\\ x_2& a_2& b_2\end{array}\right]=0$

$a_1{b}_2+b_1{x}_2+a_2{x}_1-a_1{x}_2-a_2{b}_1-b_2{x}_1=0$

$\left(a_2-b_2\right)x_1+\left(-a_1+b_1\right)x_2+\left(a_1{b}_2-a_2{b}_1\right)=0$

This is a linear equation. By computing, we can see that:

$\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{l}{a}_1\\ a_2\end{array}\right]$

And

$\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{l}{b}_1\\ b_2\end{array}\right]$ Satisfy the equation.

Therefore, $\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{l}{a}_1\\ a_2\end{array}\right]$ and $\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{l}{b}_1\\ b_2\end{array}\right]$ are a solution.