# How do you write an equation of a line through (-5,3) parallel to y=2/3x+3?

Nov 15, 2016

$y = \frac{2}{3} x + \frac{19}{3}$

#### Explanation:

Parallel lines have the same slopes. So your new line has slope $\frac{2}{3}$

Apply point-slope formula, $y - {y}_{1} = m \left(x - {x}_{1}\right)$ where $\left({x}_{1} , {y}_{1}\right)$ is $\left(- 5 , 3\right)$:
$y - 3 = \frac{2}{3} \left(x - \left(- 5\right)\right)$
$y - 3 = \frac{2}{3} \left(x + 5\right)$
$y - 3 = \frac{2}{3} x + \frac{10}{3}$
$y = \frac{2}{3} x + \frac{19}{3}$
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Nov 15, 2016

$y = \frac{2}{3} x + \frac{19}{3}$

#### Explanation:

We're going to use two things here, point-slope form and slope-intercept form. Slope intercept is $y = m x + b$, whereas point-slope is $y - {y}_{1} = m \left(x - {x}_{1}\right)$. The point that we know we go through can be written as $\left({x}_{1} , {y}_{1}\right)$ so ${x}_{1} = - 5$ and ${y}_{1} = 3$. We know the slope from the equation given in the question by applying it to slope-intercept. The $m$ is the slope, so the slope of the equation in question is $\frac{2}{3}$.

Just to organize this better:
${x}_{1} = - 5$
${y}_{1} = 3$
$m = \frac{2}{3}$

Now we plug this information into point-slope $y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - 3 = \frac{2}{3} \left(x - \left(- 5\right)\right) \to y - 3 = \frac{2}{3} \left(x + 5\right)$

This is a good answer but I'm just going to bring it to y intercept because that's what I'm used to personally.

Add $3$ to both sides.

$y = \frac{2}{3} \left(x + 5\right) + 3$

Distribute $\frac{2}{3}$

$y = \frac{2}{3} x + \frac{10}{3} + 3$

Add $\frac{10}{3}$ and $3$

$y = \frac{2}{3} x + \frac{19}{3}$