Chapter 1.1, Problem 64E

To determine

the equation of the line that passes through the point $\left(-8,0.6\right)\text{and }\left(2,-2.4\right)$ using slope intercept form if possible and draw graph.

Answer to Problem 64E

  $3x+10y+18=0$

Explanation of Solution

Given:

Given points are $\left(-8,0.6\right)\text{and }\left(2,-2.4\right)$ .

Calculation:

Standard equation for slope-intercept form is.

  $y=mx+b$

Here, $m$ is the slope of the line and $b$ is the y -intercept of the line.

Slope is calculated as.

  $m=\frac{y_2-y_1}{x_2-x_1}$

Substituting $x_1=-8,y_1=0.6,x_2=2\text{and }{y}_2=-2.4$ .

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{-2.4-0.6}{2-\left(-8\right)}\\ m=\frac{-3}{10}\\ m=-\frac{3}{10}\end{array}$

Now, substituting $x=-8,y=0.6,m=-\frac{3}{10}$ in equation $y=mx+b$ .

  $\begin{array}{l}y=mx+b\\ 0.6=\left(-\frac{3}{10}\right)\left(-8\right)+b\\ 0.6=\frac{24}{10}+b\\ 0.6=2.4+b\\ b=-1.8\end{array}$

Now, substituting $m=-\frac{3}{10}\text{ and }b=-1.8$ in equation $y=mx+b$ .

  $\begin{array}{l}y=mx+b\\ y=\left(-\frac{3}{10}\right)x+\left(-1.8\right)\\ 10y=-3x-18\\ 3x+10y+18=0\end{array}$

Graph is drawn for coordinates as.

  $\begin{array}{cc}x& y\\ 0& -1.8\\ -6& 0\end{array}$

  

As it is clearly visible from the graph that it is straight line passing through $y=-1.8$ and $x=-6$ .

Therefore, the answer is $3x+10y+18=0$ .