Q. 57 - Precalculus Enhanced with Graphing Utilities

Question

Geometry: Equation of a Line An equation of the line containing the two points $\left(x_1,y_1\right) \text{and } \left(x_2,y_2\right)$ may be expressed as the determinant

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

Prove this result by expanding the determinant and comparing the result to the two-point form of the equation of a line.

Step-by-Step Solution

Answer

The result is proven that the simplified expression is equal to the two-point form of the equation of a line..

Step 1. Given Information

We are given, 

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

We have to Prove the given result by expanding the determinant and comparing the result to the two-point form of the equation of a line 

The two-point form of the equation of a line is given by,

$\left(y-y_1\right)=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}·\left(x-x_1\right)$

Step 2. Expanding the determinant

Expanding the determinant, we get

$$\begin{gathered} \left|\begin{array}{lll}x& y& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|=0 \\ (-1{)}^{1+1}x\left|\begin{array}{ll}{y}_1& 1\\ y_2& 1\end{array}\right|+(-1{)}^{1+2}y\left|\begin{array}{ll}{x}_1& 1\\ x_2& 1\end{array}\right|+(-1{)}^{1+3}(1)\left|\begin{array}{ll}{x}_1& y_1\\ x_2& y_2\end{array}\right|=0 \\ x\left(y_1-y_2\right)-y\left(x_1-x_2\right)+1\left(x_1{y}_2-x_2{y}_1\right)=0 \\ x\left(y_1-y_2\right)+y\left(x_2-x_1\right)+x_1{y}_2-x_2{y}_1=0 \end{gathered}$$

Adding $\left(x_2{y}_1-x_1{y}_2\right)$ on both sides of the equation, we get

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

Step 3. Simplifying the expression

Subtracting $\left(x_2-x_1\right)y_1+\left(y_1-y_2\right)x$ on both sides of the equation, we get

$$\begin{gathered} x\left(y_1-y_2\right)+y\left(x_2-x_1\right)-\left[\left(x_2-x_1\right)y_1+\left(y_1-y_2\right)x\right]=x_2{y}_1-x_1{y}_2-\left[\left(x_2-x_1\right)y_1+\left(y_1-y_2\right)x\right] \\ y\left(x_2-x_1\right)-\left(x_2-x_1\right)y_1=x_2{y}_1-x_1{y}_2-\left(x_2-x_1\right)y_1-\left(y_1-y_2\right)x \end{gathered}$$

Simplifying the expression, 

$$\begin{gathered} \left(x_2-x_1\right)\left(y-y_1\right)=-\left(y_1-y_2\right)x+x_1\left(y_1-y_2\right) \\ \left(x_2-x_1\right)\left(y-y_1\right)=\left(y_1-y_2\right)\left(x_1-x\right) \\ \left(x_2-x_1\right)\left(y-y_1\right)=\left(y_2-y_1\right)\left(x-x_1\right) \end{gathered}$$

Dividing by $\left(x_2-x_1\right)$ on both sides,

$$\begin{gathered} \frac{\left(x_2-x_1\right)\left(y-y_1\right)}{\left(x_2-x_1\right)}=\frac{\left(y_2-y_1\right)\left(x-x_1\right)}{\left(x_2-x_1\right)} \\ \left(y-y_1\right)=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}·\left(x-x_1\right) \end{gathered}$$

This is the point slope form of the equation of a line.

Hence Proved.