Chapter 3, Problem 17CLR

To determine

To calculate: To find an equation of a line in standard from.

Answer to Problem 17CLR

The equation of the line in standard form is $7x-2y=-22$ .

Explanation of Solution

Given information: The line passes through the points $\left(-2,4\right)\text{ and }\left(-3,\frac{1}{2}\right)$ .

Formula used: The slope of the line passing through the points $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is given by,

  $m=\frac{y_2-y_1}{x_2-x_1}$

• Here, m is the slope of the line.

The equation of line passing through the point $\left(x_1,y_1\right)$ and having slope m in standard form is,

  $y-y_1=m\left(x-x_1\right)$

• Here, m is the slope of the line.

Calculation: It is given that the line passes through the points $\left(-2,4\right)\text{ and }\left(-3,\frac{1}{2}\right)$ . Comparing the points $\left(-2,4\right)\text{ and }\left(-3,\frac{1}{2}\right)$ with $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ ,

  $\begin{array}{c}{x}_1=-2\\ y_1=4\\ x_2=-3\\ y_2=\frac{1}{2}\end{array}$

Substitute these values in $m=\frac{y_2-y_1}{x_2-x_1}$ to find the slope of the equation,

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{\frac{1}{2}-4}{-3-\left(-2\right)}\\ =\frac{\frac{1-8}{2}}{-3+2}\\ =\frac{-7}{-2}\\ =\frac{7}{2}\end{array}$

Substitute these values in $y-y_1=m\left(x-x_1\right)$ ,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-4=\frac{7}{2}\left(x-\left(-2\right)\right)\\ y-4=\frac{7}{2}\left(x+2\right)\\ 2y-8=7x+14\\ 7x-2y=-8-14\\ 7x-2y=-22\end{array}$

Hence, the equation of the line in standard form is $7x-2y=-22$ .