# help with slopes!

• May 3rd 2007, 05:49 AM
alwaysalillost
help with slopes!
Find the slope of a line between the points (5, 3) and (-2, -3).

Write an equation for the following problems:
A line with slope 1/3 and y-intercept 4
A line with slope 3 going through the point (1, -2)
A line going through the points (3, 2) and (-2, 4)

I am totally confused on how to find the answers to these types of problems. An explaination would be great. Thankies!
• May 3rd 2007, 06:32 AM
janvdl
Always remember the formula:
y = mx + c
Where m is the slope or gradient. And c is the y-intercept

------------------------------------------------------------

1. y = 1/3x + 4 [Here we only needed to insert the values]

-------------------------------------------------------------

2. Ok, this one is a little more tricky because we have to solve for the c-value.

So: y = 3x + c

But they give us the point (1 ; -2)

So insert that into the x and y values.

Then -2 = 3(1) + c
and c = -5

Thus y = 3x - 5

----------------------------------------

3. Now we have no values. Only points.

But you know that m = (y2-y1)/(x2-x1)
So then m = (4 - 2)/(-2 - 3)
Thus m = 2/-5
Now we once again have to solve for c like in the 2nd problem.

So insert one of the given points into the equation.

then 2 = (2/-5)(3) + c
So c = 5,6

Thus y = (2/-5)x + 5,6

Hmm, thats a weird value... :eek:

Anyway i hope this helps :)
• May 3rd 2007, 12:14 PM
alwaysalillost
hmm...I'm still a little confused...:confused:
• May 3rd 2007, 01:19 PM
ecMathGeek
Quote:

Originally Posted by alwaysalillost
hmm...I'm still a little confused...:confused:

Can you be more specific on what's confusing you?
• May 3rd 2007, 01:45 PM
alwaysalillost
Quote:

Originally Posted by alwaysalillost
1. Find the slope of a line between the points (5, 3) and (-2, -3).

2. Write an equation for the following problems:
a) A line with slope 1/3 and y-intercept 4
b) A line with slope 3 going through the point (1, -2)
c) A line going through the points (3, 2) and (-2, 4)

Quote:

Originally Posted by janvdl

1. y = 1/3x + 4 [Here we only needed to insert the values]

-------------------------------------------------------------

2. Ok this one is a little more tricky because we have to solve for the c-value.

So: y = 3x + c

But they give us the point (1 ; -2)

So insert that into the x and y values.

Then -2 = 3(1) + c
and c = -5

Thus y = 3x - 5
----------------------------------------
3. Now we have no values. Only points.

But you know that m = (y2-y1)/(x2-x1)
So then m = (4 - 2)/(-2 - 3)
Thus m = 2/-5
Now we once again have to solve for c like in the 2nd problem.

So insert one of the given points into the equation.

then 2 = (2/-5)(3) + c
So c = 5,6

Thus y = (2/-5)x + 5,6

Hmm, thats a weird value...

Ok well can you explain to me how you get the equasions. I think that once I fully understand how to do that I'll be able to understand the problem. Thanks!
• May 3rd 2007, 02:13 PM
ecMathGeek
I'm sure you have seen this equation a lot, but you may not understand how it is used:

The "slope intercept" form of the equation of a line is:
y = mx + b

You can always use this equation to write the equation of the line. This form is especially useful if you already know the slope of the line and the y intercept.

m is the slope of the line. If you are given the slope of the line (or if you know enough information about the line to figure out what the slope is), you can just plug it into this. For example, if I know that the slope of a line is 1/2, then the equation would become: y = 1/2*x + b

b is the y intercept of the line. If you are told what the y intercept is you can just put that number here. However, sometimes you won't be told what the y intercept is, so you will have to figure out what it is. If, for example, you are told what the slope of the line is and you are given a point on the line, you can use this to find the y intercept (and since you already have the slope of the line, once you find the y intercept you can easily write the equation of the line). For example, if you are told the slope of the line is 1/2 and that the y intercept is 3, then you can write the equation of the line as: y = 1/2*x + 3

Sometimes you will have to figure out what m or b (or both) are.

If you are given two points on the line but they don't tell you the y-intercept or the slope, you have to figure out what both are.

To figure out what the slope you, use the equation for the slope:
m = (y2 - y1)/(x2 - x1) where x1, y1, x2, and y2 are the x and y coordinates of the points we are given. Normally the points you will be given will be in the form (x1,y1) and (x2,y2) (bascially, if you are given two points, the x and y coordinates of the first point are x1 and y1, the x and y coordinates of the second point are x2 and y2). Just take the coordinates you are given and plug them into the "m" equation. Once you have m, just plug it into the "slope intercept form" of the equation.

To figure out what the y intercept of the equation is, you need to already have the slope (which you would get from the step above). Once you have this and you have written the slope intercept form of the equation with the slope plugged in, then take the point you are given (or chose one of the points you are given if you have 2 or more points to choose from) and "plug" that point into the equation. Basically, what you need to do is plug the x coordinate of your point into the "x" in the equation and plug the y coordinate of your point into the "y" in the equation, then finally use algebra to slove for "b". This shouldn't be too hard to do and once you are done (and you have found what b equals), plug b back into the "slope intercept" form of the equation and you will have the equation of your line.

Is that clear?
• May 3rd 2007, 02:47 PM
alwaysalillost
Thanks so much ecMathGeek! I now understand how to do the equasions. Your eplaination was very clear. Thanks again for your help!
• May 3rd 2007, 02:59 PM
ecMathGeek
Quote:

Originally Posted by alwaysalillost
Thanks so much ecMathGeek! I now understand how to do the equasions. Your eplaination was very clear. Thanks again for your help!

I'm glad I could help. :)