Chapter 4.2, Problem 26PPS

To determine

To find: The equation of the straight line having slope $\left(m\right)=\frac{1}{3}$ and passes through the point $\left(3,-2\right)$ .

Answer to Problem 26PPS

The equation of straight line in slope-intercept form is $y=\frac{1}{2}x$ .

Explanation of Solution

Given information: The given line passes through the point $\left(3,-2\right)$ and has slope $\left(m\right)=\frac{1}{3}$ .

Formula used: Using slope-point form of the line, the equation of the line is written as $y-y_1=m\left(x-x_1\right)$ where $\left(m\right)$ is the slope of the given line and $\left(x_1,y_1\right)$ is the point through which the line passes.

Calculation: Using slope-point form of the line, the equation of the line is evaluated as

  $\begin{array}{c}y-\left(-2\right)=\frac{1}{3}\left(x-3\right)\mathrm{......}\left(\begin{array}{l}\text{Plugging for }m=\frac{1}{3}\\ \text{and }\left(x_1,y_1\right)\to \left(3,-2\right)\end{array}\right)\\ y+2=\frac{1}{3}\left(x-3\right)\mathrm{......}\left(\text{Simplifying}\right)\\ 3y+6=x-\mathrm{3......}\left(\text{Cross-multiplying}\right)\\ \text{or, }x-3y-9=0\end{array}$

Hence, the required equation of the straight linehaving slope $\left(m\right)=\frac{1}{3}$ and passes through the point $\left(3,-2\right)$ is $x-3y-9=0$ .