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Algebra 20 Online
OpenStudy (anonymous):

How to graph ( -3, -1) and the slope is m= -1/5 please show me in details with explanation

OpenStudy (anonymous):

m=x/y just it !:)

OpenStudy (anonymous):

you want the equation of the line?

OpenStudy (anonymous):

Nope

OpenStudy (anonymous):

so what is it? :D

OpenStudy (anonymous):

Never mind I got it thank you

OpenStudy (anonymous):

Nope

OpenStudy (anonymous):

sorry I think m=x/y is discorrect!

OpenStudy (anonymous):

say what do you want

OpenStudy (anonymous):

I think I should graph the points then I take the slope either I go up or move right

OpenStudy (anonymous):

Am I right?

OpenStudy (anonymous):

yes you're right but you should use this formula: tan(Theta) = m

OpenStudy (anonymous):

I m sure for this :)

OpenStudy (anonymous):

there is no problem :)

OpenStudy (anonymous):

x/y is incorrect,(y2-y1)/(x2-x1) is correct and then you can use this: y/x not x/y!

OpenStudy (anonymous):

and besides that,line pasts two points not one points!

OpenStudy (anonymous):

Why should I use tan (theta) =m

OpenStudy (anonymous):

I take algebra 2 only

OpenStudy (anonymous):

if you have "mesure of angle" you should use tan(theta) but if you know "2 points" you can use (y2-y1)/(x2-x1)

OpenStudy (anonymous):

K I got u thnx!

OpenStudy (anonymous):

You're welcome,Have a look at this too: http://en.wikipedia.org/wiki/Slope

OpenStudy (anonymous):

got it?

OpenStudy (anonymous):

Yup sure did :) thnx again

OpenStudy (anonymous):

You're welcome:)

OpenStudy (debbieg):

No need for trig in this problem. You are on the right track with plotting the point and then using the slope to get a 2nd point, but I worry what you mean by "either I go up or move right". :) You need to use the slope to determine the RISE and the RUN. You will EITHER move up or down the RISE, and you will EITHER move left or right the RUN. The combination depends on the sign of the slope. For positive slope, you want x and y to change in the same direction.... so you either go UP and RIGHT (both y and x increase) or DOWN and LEFT (both y and x decrease), in the amounts, respectively, of the rise and run. Either will work and will give you a 2nd points. For negative slope, you want x and y to change in opposite directions.... so you either go UP and LEFT, or DOWN and RIGHT. Once you have that 2nd point, draw the line through those 2 points. This method will work fine, although it isn't quite as "slick" as just cooking up the equation for the line. Then you can get any 2nd point on the line from the equation (like say, the y-intercept).

OpenStudy (anonymous):

Wow!!! Ur good thanx!!! But what if my slope is negative like -3 should I do this -3/1 then find it?

OpenStudy (debbieg):

Yes, you can interpret that slope as: \(\Large m=-3=-\dfrac{3}{1}=-\dfrac{rise}{run}\) So for each change of 3 units of "rise" (in the vertical direction, or in the y-coordinate, however you find it most natural to think of it), you will change 1 unit in "run" (in the horizontal direction, or in the x-coordinate). AND since slope is NEGATIVE, the changes must go in OPPOSITE directions. So from wherever your starting point is, you could either go: Up 3 and left 1 (increasing y, decreasing x) or Down 3 and right 1 (decreasing y, increasing x) Either works fine, they just give you two different options for the "2nd point".

OpenStudy (anonymous):

OMG!!!! THANK YOOOUUU!!!!! I REALLY GET IT!!!

OpenStudy (anonymous):

Could the second point be 3, -1?

OpenStudy (anonymous):

@DebbieG

OpenStudy (anonymous):

Could the second point be 3, -1?

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