Chapter 12.7, Problem 8MRE

To determine

To Calculate: An equation, in standard form, of the line passing through the given points.

  $\left(3,2\right),\left(-6,-4\right)$

Answer to Problem 8MRE

The equation of the line in standard form is $2x-3y=0$

Explanation of Solution

Equation of line is standard form is

  $ax+by=c$...... (1)

And the equation of line passing through two points is represent

  $\frac{y-y_1^{}}{y2-y_1^{}}=\frac{x-x_1^{}}{x_2^{}-x_1^{}}$

Where $\left(x_{1,}^{}{y}_1^{}\right)$ and $\left(x_{2,}^{}{y}_2^{}\right)$ are the points through which line is passing.

Given points are:

  $\left(3,2\right),\left(-6,-4\right)$

Put the points in equation 2.

  $\begin{array}{l}\frac{y-2}{-4-2}=\frac{x-3}{-6-3}\\ \frac{y-2}{-6}=\frac{x-3}{-9}\end{array}$

Cancel negative at both sides:

  $\frac{y-2}{6}=\frac{x-3}{9}$

  $\frac{y-2}{2}=\frac{x-3}{3}$

  $3\left(y-2\right)=2\left(x-3\right)$

  $3y-6=2x-6$

  $3y=2x-6+6$

  $3y=2x$

So the equation of the line in standard form is given below which is in the same form as equation 1.

  $2x-3y=0$