Chapter 4.3, Problem 47PPS

To determine

To find: The equation of the straight line $\left(l\right)$ in point-slope form, slope-intercept form and standard form which passes through the point $\left(-9,4\right)$ having slope $\left(m\right)=\frac{4}{7}$ .

Answer to Problem 47PPS

The equation of straight line $\left(l\right)$ in point-slope form $y-4=\frac{4}{7}\left(x+9\right)$ , slope-intercept form $y=\frac{4}{7}x+\frac{64}{7}$ and standard form is $\text{4}x-7y=-64$ .

Explanation of Solution

Given information: The given line $\left(l\right)$ has slope $\left(m\right)=\frac{4}{7}$ and it passes through the point $\left(-9,4\right)$ .

Calculation: The equation of the straight line $\left(l\right)$ in point-slope form having slope $\left(m\right)=\frac{4}{7}$ and passing through the point $\left(-9,4\right)$ is written as,

  $\begin{array}{c}y-4=\frac{4}{7}\left(x-\left(-9\right)\right)\mathrm{......}\left(\begin{array}{l}\text{Using point slope form}\\ y-y_1=m\left(x-x_1\right),m=\frac{4}{7}\\ \text{and }\left(x_1,y_1\right)\to \left(-9,4\right)\end{array}\right)\\ y-4=\frac{4}{7}\left(x+9\right)\mathrm{......}\left(1\right)\end{array}$

For slope-intercept form the equation of line $\left(l\right)$ in slope-intercept form can be obtained as,

  $\begin{array}{c}y-4=\frac{4}{7}\left(x+9\right)\mathrm{......}\left(\begin{array}{l}\text{Using slope-intercept form}\\ y=mx+c\text{ and using }\left(1\right)\end{array}\right)\\ y-4=\frac{4}{7}x+\frac{36}{7}\\ y=\frac{4}{7}x+\frac{64}{7}\mathrm{......}\left(\text{Simplifying}\right)\end{array}$

Similarly, the standard form of line $\left(l\right)$ from (1) can be obtained as,

  $\begin{array}{c}y-4=\frac{4}{7}\left(x+9\right)\mathrm{......}\left(\begin{array}{l}\text{Using standard form,}\\ ax+by=c\text{ and using }\left(1\right)\end{array}\right)\\ 7\left(y-4\right)=4\left(x+9\right)\mathrm{......}\left(\text{Cross-multiplying}\right)\\ 7y-28=4x+36\\ \text{or, 4}x-7y=-64\end{array}$