How to write equation in slope-intercept form of indicated line that goes through (5, -6) and (2, -8)?

Question

So I have this math problem on my review packet it is extremely important worth 50 points I have an A in the class but don't understand what this problem is asking can you help me solve it/explain it to me?

Answer 1

Hint:

I suppose that the ''slope-intercept'' form of the equation of a line is $y=mx+q$. You have two points: substitute the coordinates of the two points and find $m$ and $q$ solving a linear system.

Answer 2

You can use the point-slope equation.

y - (y coordinate) = slope * (x - (x coordinate))

Just plug in the x and y coordinates and rearrange the terms to get a slope-intercept equation.

Hope that helps!

Answer 3

The following explanation if just for students that want to understand the concept behind the algebra. It also makes the problem a lot easier to think about.

Let's start with a simple set of points (1,3), (2, 7)

Put these guys in a table and extend the pattern
$\begin{matrix} x & y \\ 1 & 3 \\ 2 & 7 \\ 3 & 11 \end{matrix}$

The question then becomes, "How do we change the list of x's into the list of y's?".

Noticing that the list of y's goes up by adding 4 each time, we multiply our list of x's by 4.

$\begin{matrix} x & 4x \\ 1 & 4 \\ 2 & 8 \\ 3 & 12 \end{matrix}$

Then we write our goal next to the column we just made.

$\begin{matrix} x & 4x & 4x-1\\ 1 & 4 & 3\\ 2 & 8 & 7\\ 3 & 12 & 11\end{matrix}$

And we see that we just have to subtract 1 from our middle list to get to our list of y's. So

$y=4x-1$.

Next, let's look at a case with the points (2, 3), (4, 6)

$\begin{matrix} x & y \\ 2 & 3 \\ 4 & 6 \\ 6 & 9 \end{matrix}$

We notice the x list goes up by 2's and the y list goes up by 3's.

So we set up our table.

$\begin{matrix} x & \frac{x}{2} & \frac{3 x}{2} \\ 2 & 1 & 3 \\ 4 & 2 & 6 \\ 6 & 3 & 9 \end{matrix}$

So here, $y=\frac{3}{2} x $

Your problem is like a combination of the two problems I've just run. See if you can understand the process and it will help out not only with this problem, but with a ton of other problems in your algebra course.

So here's a problem sort of like yours. If we had points (1, 5), (3, 8)

$\begin{matrix} x & y \\ 1 & 5 \\ 3 & 8 \\ 5 & 11 \end{matrix}$

We start by just working on the x's. We change the list to (just about ) the simplest list possible, 0, 1, 2.

$\begin{matrix} x & x-1 & \frac{x-1}{2} \\ 1 & 0 & 0 \\ 3 & 2 & 1 \\ 5 & 4 & 2 \end{matrix}$

Then we take a look at the y's, and do the same sort of process.

$\begin{matrix} y & y-5 & \frac{y-5}{3} \\ 5 & 0 & 0 \\ 8 & 3 & 1 \\ 11 & 6 & 2 \end{matrix}$

Since our two tables here have the same last column, those terms are equal to each other.

$\frac{x-1}{2}=\frac{y-5}{3}$

This is called 'symmetric form'. Which can easily be changed to point slope form.

$y-5=\frac{3}{2} (x-1)$

And then to slope intercept form.

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

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

$y=\frac{3}{2} x+\frac{7}{2}$