Chapter 4.3, Problem 46PPS

To determine

To find: The equation of the straight line in point-slope form, slope-intercept form andstandard form which passes through the point $\left(-2,8\right)$ having slope $\left(m\right)=\frac{8}{5}$ .

Answer to Problem 46PPS

The equation of straight line in point-slope form $y-8=\frac{8}{5}\left(x+2\right)$ , slope-intercept form $y=\frac{8}{5}x+\frac{56}{5}$ and standard form is $8x-5y=-56$ .

Explanation of Solution

Given information: The given line has slope $\left(m\right)=\frac{8}{5}$ and it passes through the point $\left(-2,8\right)$ .

Calculation: The equation of the straight line in point-slope form having slope $\left(m\right)=\frac{8}{5}$ and passing through the point $\left(-2,8\right)$ is written as,

  $\begin{array}{c}y-8=\frac{8}{5}\left(x-\left(-2\right)\right)\mathrm{......}\left(\begin{array}{l}\text{Using point slope form}\\ y-y_1=m\left(x-x_1\right),m=\frac{8}{5}\\ \text{and }\left(x_1,y_1\right)\to \left(-2,8\right)\end{array}\right)\\ y-8=\frac{8}{5}\left(x+2\right)\mathrm{......}\left(1\right)\end{array}$

For slope-intercept form the equation of line in slope-intercept form can be obtained as,

  $\begin{array}{c}y-8=\frac{8}{5}\left(x+2\right)\mathrm{......}\left(\begin{array}{l}\text{Using slope-intercept form}\\ y=mx+c\text{ and using }\left(1\right)\end{array}\right)\\ y-8=\frac{8}{5}x+\frac{16}{5}\\ y=\frac{8}{5}x+\frac{56}{5}\mathrm{......}\left(\text{Slope-intercept form}\right)\end{array}$

Similarly, the standard form of line from (1) can be obtained as,

  $\begin{array}{c}y-8=\frac{8}{5}\left(x+2\right)\mathrm{......}\left(\begin{array}{l}\text{Using standard form,}\\ ax+by=c\text{ and using }\left(1\right)\end{array}\right)\\ 5\left(y-8\right)=8\left(x+2\right)\mathrm{......}\left(\text{Cross-multiplying}\right)\\ 5y-40=8x+16\\ \text{or, }8x-5y=-56\end{array}$