Chapter 11.3, Problem 57AYU

Geometry: Equation of a inline An equation of the inline containing the two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ may be expressed as the determinant $\left[\begin{array}{c}x\\ x_1\\ x_2\end{array}\begin{array}{c}y\\ y_1\\ y_2\end{array}\begin{array}{c}1\\ 1\\ 1\end{array}\right]=0$ Prove this result by expanding the determinant and comparing the result to the two-point form of the equation of a inline

To determine

To find: The result to the two-point form of the equation of a line is same as the expanded form of the determinant equation:

$\left|\begin{array}{ccc}x& y& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|=0$

Answer to Problem 57AYU

$x{y}_1-x{y}_2-y{x}_1+y{x}_2+x_1{y}_2-x_2{y}_1=0$

Explanation of Solution

Expanding the following determinant equation we get,

$\left|\begin{array}{ccc}x& y& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|=0$

$x\left(y_1-y_2\right)-y\left(x_1-x_2\right)+1\left(x_1{y}_2-x_2{y}_1\right)=0$

$x{y}_1-x{y}_2-y{x}_1+y{x}_2+x_1{y}_2-x_2{y}_1=0$

Moreover we know that an equation of a straight line passing through two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is:

$\frac{x-x_1}{x_1-x_2}=\frac{y-y_1}{y_1-y_2}$

$\left(x-x_1\right)\left(y_1-y_2\right)=\left(y-y_1\right)\left(x_1-x_2\right)$

$x{y}_1-x_1{y}_1-x{y}_2+x_1{y}_2=y{x}_1-x_1{y}_1-y{x}_2+y_1{x}_2$

$x{y}_1-x_1{y}_1-x{y}_2+x_1{y}_2-y{x}_1+x_1{y}_1+y{x}_2-y_1{x}_2=0$

$x{y}_1-x{y}_2-y{x}_1+y{x}_2+x_1{y}_2-x_2{y}_1=0$

The above is same as the expanded form of determinant equation.

Hence the proof.