Niamh D.
asked 08/29/17Determine the equation of the line that passes through the points (- 3, - 2) and (0, 0). Use the form y = mx + b.
Determine the equation of the line that passes through the points (- 3, - 2) and (0, 0). Use the form y = mx + b.
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2 Answers By Expert Tutors
Ellen M. answered 08/29/17
Tutor
4.9
(705)
NASA Math/Physics Tutor - 20 Years Experience
Hi guys,
For y = mx + b, m is the slope, because that's the number that says how y is affected when you change x. b is just a constant, but it does tell you what y is when x = 0, because at that point it doesn't matter what m is, anything times zero is just zero and all you're left with is the b.
So the most important and easiest piece of information to find first is the slope.
(Without knowing how x is changing y in the first place, it would be pretty hard to figure out what it's doing when x = 0!)
Slope is defined as "rise over run", because it's telling you how the rise is changing with a given amount of run, (usually 1). But you can use any two points to measure, because fractions can always reduce if needed. In this case we have (-3, -2) and (0, 0). "Rise" is literally how far you rose, or how much you increased your y.
If I had the following points: (2,1) and (4,2)
-
| (4,2) <y = 2
- x
| (2,1) | <Rise = 2 - 1 = 1. (It rose a distance of 1,)
- x __ __ <y = 1 from 1 to 2
|
---|----|----|----|----|----|--------|----|----|----|----|----|---------
|
| ^ ^
| x = 1 x = 3
|
| Run = 3 - 1 = 2. (It ran a distance of 2, from 1 to 3).
|
| ^ ^
| x = 1 x = 3
|
| Run = 3 - 1 = 2. (It ran a distance of 2, from 1 to 3).
|
So in this case, our slope would be Rise/Run, i.e 1/2. We could then use that as the value for m and solve for b.
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Carol H. answered 08/29/17
Tutor
4.9
(285)
Experienced Mathematics Tutor w/ Master's Degree in Math
The slope of the line is m = (y2-y1)/(x2-x1)
m = (-2-0)/-3-0) = -2/-3 = 2/3
The y=intercept is 0.
y = 2/3 x + 0
y = 2/3 x
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Ellen M.
Rise is y2 - y1, (the distance between the two y's).
Run is x2 - x1, (the distance between the two x's).
And altogether, slope is Rise/Run, or (y2 - y1)/(x2 - x1).
So in the original problem, our points are (-3, -2), and (0, 0). (It doesn't matter which point you pick to be first, just as long as you're consistant with your choice across both the x's and the y's. In this case I'll let (0,0) be the second point so it's easy to subtract. So... Rise = (-2) - (0) = -2, and Run = (-3) - (0) = -3, so m = Slope = Rise/Run = (-2)/(-3). Double negatives cancel, so m = positive 2/3.
Now that we know that m = 2/3, we can plug it into y = mx + b, so now y = 2/3 x + b. m and b are constants, they always stay the same, but x and y are variables and are subject to change. However, we can use any single snapshot (i.e. either of the two points we were given), plug in that particular x and y since we already know they go together, figure out what b is in that instance, and then know that b will always be that same number.
Again, I like making life easy for myself, so I'll choose (0,0) to plug in. (It would get you the exact same answer if you chose (-3,-2), but I'm lazy. :-D
So (0,0) means y = 0 when x = 0, and we already know m = 2/3, so I'm going to plug all that in and solve for b.
y = m * x + b turns into....
0 = 2/3 * (0) + b but zero times anything is 0, (order of operations! multiplication before addition...), so it's really
0 = 0 + b meaning that b = 0. (All that work for nothing!! :-)
So now, since b is constant, it is not changing, so once zero always zero, and we are left with the following:
m is always 2/3.
b is always 0.
x and y are subject to change, so we'll just leave them generic.
Plugging all this into y = mx + b,
You are left with y = 2/3 x + 0
Otherwise known as y = 2/3x.
We can check the original two points we were given:
(0,0) Is 0 = 2/3 * 0? Check.
(-3,-2) Is -2 = 2/3 * -3? Check.
Hooray!!! :-)
So I know I probably way overexplained this, but I wanted to make sure anyone could come along and understand what I did from start to finish, no matter their math background. Tutoring one-on-one allows the teacher to zoom in on what that student particular student needs, so you don't have to read through lots of explanation you already know, and hopefully can even have fun in the process. :-)
Happy mathing!!
-Ellen M.
08/29/17