# Question #d842e

Jan 24, 2017

$\left(y + \textcolor{red}{1}\right) = \textcolor{b l u e}{\frac{2}{3}} \left(x - \textcolor{red}{3}\right)$

Or converting to the slope-intercept form:

$y = \frac{2}{3} x - 3$

#### Explanation:

First, we need to find the slope of the equation given in the problem by converting it to the familiar slope-intercept form by solving for $y$:

$2 x - \textcolor{red}{2 x} - 3 y = - \textcolor{red}{2 x} + 8$

$0 - 3 y = - 2 x + 8$

$- 3 y = - 2 x + 8$

$\frac{- 3 y}{\textcolor{red}{- 3}} = \frac{- 2 x + 8}{\textcolor{red}{- 3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 3}}} y}{\cancel{\textcolor{red}{- 3}}} = \frac{2}{3} x - \frac{8}{3}$

$y = \frac{2}{3} x - \frac{8}{3}$

The slope-intercept form of a linear equation is:

$y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

Therefore we know the slope of this line and a line parallel to this line is $\textcolor{red}{m = \frac{2}{3}}$

We can now use this slope and the point and the point-slope formula to find an equation for the line requested in the problem:

The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$

Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.

Substituting the values from the problem and the calculate gives:

$\left(y - \textcolor{red}{- 1}\right) = \textcolor{b l u e}{\frac{2}{3}} \left(x - \textcolor{red}{3}\right)$

$\left(y + \textcolor{red}{1}\right) = \textcolor{b l u e}{\frac{2}{3}} \left(x - \textcolor{red}{3}\right)$

Or converting to the slope-intercept form:

$y + \textcolor{red}{1} = \frac{2}{3} x - \left(\frac{2}{3} \times \textcolor{red}{3}\right)$

$y + 1 = \frac{2}{3} x - 2$

$y + 1 - 1 = \frac{2}{3} x - 2 - 1$

$y = \frac{2}{3} x - 3$

Jan 24, 2017

$2 x - 3 y = 9$

#### Explanation:

$2 x - 3 y = 8$
we change to $y = m x + c$, where $m$=gradient.

$2 x - 8 = 3 y$
$y = \frac{2 x - 8}{3}$
$y = \frac{2}{3} x - \frac{8}{3}$

since it is a parallel line, they have a same value of gradient,where $m = \frac{2}{3}$.

use $y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - \left(- 1\right) = \frac{2}{3} \left(x - 3\right)$
$y + 1 = \frac{2}{3} x - 2$
$y = \frac{2}{3} x - 2 - 1$
$y = \frac{2}{3} x - 3$
$3 y = 2 x - 9$
$9 = 2 x - 3 y$