Chapter 8.6, Problem 14E

To determine

To write:

Anequation of the line from the given graph.

Answer to Problem 14E

The equation of the line passing through the given pointswould be $y=-2x+3$ .

Explanation of Solution

Given:

A graph:

Concept used:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1},\text{where,}\\ m=\text{Slope of line,}\\ y_2-y_1=\text{Difference between two }y\text{-coordinates}\\ x_2-x_1=\text{Difference between two }x\text{-coordinates of same }y-\text{coordinates}\end{array}$

The equation of a line in slope-intercept form is $y=mx+b$ , where m represents slope of line and b represents the y -intercept.

Calculation:

Upon looking at our given graph, we can see that line crosses through the points $(0,3)\text{and }(2,-1)$

Let us find the slope of line passing through the given points.

Let point $(0,3)=(x_1,y_1)\text{and }(2,-1)=(x_2,y_2)$ . Upon substituting our given coordinates in slope formula, we will get:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{-1-3}{2-0}\\ m=\frac{-4}{2}\\ m=-2\end{array}$

To write our equation, we will substitute the value of slope and the y -intercept $(0,3)$ in slope-interceptform of equation as shown below:

  $\begin{array}{l}y=mx+b\left(\text{Slope-intercept form of equation}\right)\\ y=-2\cdot x+3\left(\text{Substituting }m=-2\text{ and }b=3\right)\\ y=-2x+3\end{array}$

Therefore, the equation of the line passing through the given points would be $y=-2x+3$ .