# How do you write a line parallel to y=-3/5x+2 and goes through point (0,-2)?

Dec 4, 2014

$y = - \frac{3}{5} x - 2$

Explanation:

For parallel lines, the slope is the same. Since this given line is already given in the form $y = m x + b$, we know that the slope of any parallel line will also be $m$. Here, $m = - \frac{3}{5}$ from $y = - \frac{3}{5} x + 2$.

The question is asking us to find an equation of the line that passes through the point $\left(0 , - 2\right)$. Well, this means that our line will have a point where $x = 0$ and $y = - 2$, but with a given slope of $m = - \frac{3}{5}$. We just need to find a $b$ such that the point $\left(0 , - 2\right)$ exists on that line.

We can do this in two ways.

1) Use $y = m x + b$ with $x = 0$, $y = - 2$, and $m = - \frac{3}{5}$ to solve $b$ and find this general equation of the parallel line.

Let's plug in our values.

$- 2 = \left(- \frac{3}{5}\right) \left(0\right) + b$
$b = - 2$

We now know our $b$, which we can plug into the new $y = m x + b$ (of which we already know $m$, since it remains the same). So, the equation of the line parallel to $y = - \frac{3}{5} x + 2$ that goes through the point $\left(0 , - 2\right)$ is

$y = - \frac{3}{5} x - 2$

OR

2) Use the point-slope form of $y - {y}_{1} = m \left(x - {x}_{1}\right)$, where ${y}_{1}$ and ${x}_{1}$=coordinates of point on line and $m$=slope, to find the equation directly. Here, ${x}_{1} = 0$ and ${y}_{1} = - 2$, from $\left(0 , - 2\right)$ and $m$ is still $- \frac{3}{5}$.

Let's solve to find the equation of the line.
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - \left(- 2\right) = - \frac{3}{5} \left(x - 0\right)$
$y + 2 = - \frac{3}{5} x$
$y = - \frac{3}{5} x - 2$