Chapter 4.1, Problem 6MRE

a.

To determine

The slope of the line containing the points $\left(0,0\right),\left(-3,2\right)$ .

Answer to Problem 6MRE

The value of slope is $-\frac{2}{3}$ .

Explanation of Solution

Given information: The two points contained by line are $\left(0,0\right),\left(-3,2\right)$ .

Formula Used: The slope $\left(m\right)$ of line through two points $\left(x_1,y_1\right),\left(x_2,y_2\right)$ is given as,

  $m=\frac{y_2-y_1}{x_2-x_1}$

Calculation: The slope of the line through the points $\left(0,0\right),\left(-3,2\right)$ is evaluated using the formula as,

  $\begin{array}{c}m=\frac{2-0}{-3-0}\mathrm{....}\left(\begin{array}{l}\left(x_1,y_1\right)\to \left(0,0\right)\\ \left(x_2,y_2\right)\to \left(-3,2\right)\end{array}\right)\\ =-\frac{2}{3}\mathrm{......}\left(1\right)\end{array}$

Hence the slope of the line is $-\frac{2}{3}$ .

b.

To determine

The equation of the line in standard form through the points $\left(0,0\right),\left(-3,2\right)$ .

Answer to Problem 6MRE

The equation of line in standard form is $\text{2}x+3y=0$ .

Explanation of Solution

Given information: The two points contained by line are $\left(0,0\right),\left(-3,2\right)$ .

Formula Used: The equation of line having slope $\left(m\right)$ and through the point $\left(x_1,y_1\right)$ is given as,

  $y-y_1=m\left(x-x_1\right)$

Calculation: The equation of the line through the point $\left(0,0\right)$ and having slope $\left(m=-\frac{2}{3}\right)$ from (1) is evaluated using the formula as,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-0=\left(-\frac{2}{3}\right)\left(x-0\right)\mathrm{....}\left(\begin{array}{l}\left(x_1,y_1\right)\to \left(0,0\right)\\ m=-\frac{2}{3}\text{ from }\left(1\right)\end{array}\right)\\ 3y=-2x\mathrm{......}\left(\text{Cross-multiply}\right)\\ \text{or, 2}x+3y=\mathrm{0......}\left(\begin{array}{l}\text{Standard form}\\ ax+by=c\end{array}\right)\end{array}$