# How do you write an equation with slope of 5/3 and contains the point (-6, -2)?

Sep 24, 2015

$y = \frac{5}{3} x + 8$

#### Explanation:

To do this, we use a linear equation called point slope form. This is basically another way of writing a linear equation, like $y = m x + b$. Point slope form is as follows: $y - {y}_{1} = m \left(x - {x}_{1}\right)$. I won't go into the specifics of what this equation is or how it's derived, but I encourage you to do so. In this equation, ${y}_{1}$ and ${x}_{1}$ are points on the line $y$ and $m$ is the slope.

Here, we already have the elements: points on the line, and the slope. To solve, we just substitute these values into the equation and simplify:
$y - \left(- 2\right) = \left(\frac{5}{3}\right) \left(x - \left(- 6\right)\right)$; ${x}_{1} = - 6$, ${y}_{1} = - 2$, $m = \frac{5}{3}$
$y + 2 = \frac{5}{3} \left(x + 6\right)$
$y + 2 = \frac{5}{3} x + 10$
$y = \frac{5}{3} x + 10 - 2$
$y = \frac{5}{3} x + 8$
And there you have it - the equation of the line with slope 5/3 and passing through the point (-6,-2).