# How do you write an equation of the line that passes through (–3, –5) and (3, 0)?

Jul 6, 2017

The equation of the line is:

$y = \frac{5}{6} x - \frac{15}{6}$

#### Explanation:

The equation of the line will be in the form:

$y = m x + c$

where $m$ is the slope (gradient) and $c$ is the y-intercept.

To find the slope, we use:

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

It doesn't matter which point we decide is $\left({x}_{1} , {y}_{1}\right)$ and which we choose as (x_2,y_2, since the formula will work either way.

$m = \frac{0 - \left(- 5\right)}{3 - \left(- 3\right)} = \frac{5}{6}$

Now we can use the slope and the coordinates of one point - either will do - to find the y-intercept:

$y = m x + c$

$0 = \frac{5}{6} \left(3\right) + c = \frac{15}{6} + c$

Rearranging:

$c = 0 - \frac{15}{6} = - \frac{15}{6}$

Over all, then, the equation of the line is:

$y = \frac{5}{6} x - \frac{15}{6}$

Jul 6, 2017

The line is $y = \frac{5}{6} x - \frac{5}{2.}$

#### Explanation:

The general equation of a line is given by
$$y=mx+q$$
so we need to substitute our two points and solve the two equations that we will obtain.

First equation: the point is $\left(- 3 , - 5\right)$ it means that we have to substitute $x = - 3$ and $y = - 5$ obtaining
$$-5=-3m+q.$$
Second equation: the point is $\left(3 , 0\right)$ it means that we have to substitute $x = 3$ and $y = 0$ obtaining
$$0=3m+q.$$

From the second equation we have
$$q=-3m$$
that we can substitute in the first equation obtaining
$$-5=-3m-3m$$
$$-5=-6m$$
$$5=6m$$
$$m=5/6$$
and, consequently
$$q=-3m=-3\times5/6=-5/2.$$
So the equation of the line is
$$y=5/6x-5/2.$$

To be sure that the line is correct we can substitute the two points and see that we obtain the identities. First point $\left(- 3 , - 5\right)$

$$-5=-3\times5/6-5/2$$
$$-5=-5/2-5/2$$
$$-5=-5.$$

Second point $\left(3 , 0\right)$

$$0=3\times 5/6-5/2$$
$$0=5/2-5/2$$
$$0=0.$$