# Math Help - How to find the y-intercept and slope.

1. ## How to find the y-intercept and slope.

Hi ppl.

If Im given the coordinates below:

x: 32, 48, 64, 80, 96, 112
y: 6, 9, 12, 15, 18, 21

How do I find the slope and the y-intercept of the linear equation without using regression or any type of computer software (solve algebraically)?

slope: pick to random coordinates like (32, 6) and (112, 21) and then put it into y2 - y1 / x2 - x1?

y-intercept: well.. let x be 0 in the equation?? well I cant use regression? how do I get the equation?

2. Originally Posted by djvollo
Hi ppl.

If Im given the coordinates below:

x: 32, 48, 64, 80, 96, 112
y: 6, 9, 12, 15, 18, 21

How do I find the slope and the y-intercept of the linear equation without using regression or any type of computer software (solve algebraically)?

slope: pick to random coordinates like (32, 6) and (112, 21) and then put it into y2 - y1 / x2 - x1?

y-intercept: well.. let x be 0 in the equation?? well I cant use regression? how do I get the equation?
Assuming that the above relation does, in fact, give a line your slope can be computed by picking any two points and put it through the slope formula, as you suggested.

Now we have the slope-intercept form for a line:
$y = mx + b$
where the y-intercept is the point (0, b). Pick an arbitrary point from your set and plug it into this equation. Solve for b.

-Dan

3. So e.g.: (21-6)/(112-32) = 0.1875

m = 0.1875..

then plug in a random x point into the equation.

Like y = 0.1875(6) + b??

Don't get what u mean really... arbitrary point?

Next thing is: x: 32, 48, 64, 80, 96, 112 and y: 6, 14, 24, 38, 55, 75..

How do I develop this function algebraically (it is not linear)?

4. First: when Dan suggested selecting an arbitrary point, he specifically mentioned that it needed to be from your set of points. For example, your first point is (32, 6). If you plug this point into what you have so far, you get:

$6 = 0.1875 (32) + b$.
You'll find that, regardless of which point in your set you select, you'll get the same b.

For your new problem, it sounds like you might be doing your first problems of parametrics. The basic idea is to describe your x's and y's in terms of a third variable (usually t). Then, if you want an equation relating just x and y, you often can solve one of your relations for t, and then substitue it into the other. Using your first problem as an example:

$x = 16(t + 1)$
and:

$y = 3t + 3$
$y-3 = 3t$
$\frac{y-3}{3} = t$

Then substitute that back into the first equation to get:
$x = 16(\frac{y-3}{3} + 1)$
I'll leave that as x defined in terms of y, but solving for y (or even substituting the other way) would give you the same linear equation you're looking for.