Question

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

Answer 1

The equation of the line is:

`y = 5/6x-15/6`

Explanation:

The equation of the line will be in the form:

`y = mx+c`

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

To find the slope, we use:

`m=(y_2-y_1)/(x_2-x_1)`

It doesn't matter which point we decide is `(x_1,y_1)` and which we choose as `(x_2,y_2`, since the formula will work either way.

`m=(0-(-5))/(3-(-3))=5/6`

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

`y = mx+c`

`0 = 5/6(3)+c=15/6+c`

Rearranging:

`c=0-15/6=-15/6`

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

`y = 5/6x-15/6`

Answer 2

The line is `y=5/6x-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 `(-3,-5)` it means that we have to substitute `x=-3` and `y=-5` obtaining
$$-5=-3m+q.$$
Second equation: the point is `(3,0)` 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 `(-3,-5)`

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

Second point `(3,0)`

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