# Question #bccac

Dec 3, 2017

By converting the equation in point-slope form to slope-intercept form, we obtain $y = \setminus \frac{8}{9} x + \setminus \frac{8}{3}$

#### Explanation:

Given the slope of a line and a point on it, we can write its equation in point-slope form, denoted by:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

Where $m$ is the slope, and $\left({x}_{1} , {y}_{1}\right)$ is a point on the line.

Let’s plug those values in:

$y - \left(- 8\right) = \setminus \frac{8}{9} \left(x - 6\right)$

$\setminus \implies y + 8 = \setminus \frac{8}{9} \left(x - 6\right)$

We can now move around terms to convert this to slope-intercept form:

$y = 8 = \setminus \frac{8}{9} \left(x - 6\right)$

$\setminus \implies y + 8 = \setminus \frac{8}{9} x - \setminus \frac{16}{3}$

$\setminus \implies y = \setminus \frac{8}{9} x + \setminus \frac{8}{3}$

Dec 3, 2017

$y = \frac{8}{9} x - \frac{40}{3}$

#### Explanation:

$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = m x + b} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where m is the slope and b the y-intercept}$

$\text{here } m = \frac{8}{9}$

$\Rightarrow y = \frac{8}{9} x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$

$\text{to find b substitute "(6,-8)" into the partial equation}$

$- 8 = \left(\frac{8}{9} \times 6\right) + b \Rightarrow b = - 8 - \frac{16}{3} = - \frac{40}{3}$

$\Rightarrow y = \frac{8}{9} x - \frac{40}{3} \leftarrow \textcolor{red}{\text{in slope-interceot form}}$