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Mathematics 20 Online
OpenStudy (anonymous):

Last question! @jim_thompson5910 What is the slope of a line that passes through the point (−5, 3) and is parallel to a line that passes through (2, 13) and (−4, −11)?

jimthompson5910 (jim_thompson5910):

your first task is to find the slope of the line through (2, 13) and (−4, −11

jimthompson5910 (jim_thompson5910):

(2, 13) and (−4, −11) *

OpenStudy (anonymous):

Hooooww do I do that. I forgot. o.o heh.

jimthompson5910 (jim_thompson5910):

that's ok, you use the slope formula m = (y2-y1)/(x2-x1)

jimthompson5910 (jim_thompson5910):

the first point is (x1,y1) the second point is (x2,y2)

OpenStudy (anonymous):

Ohhhhh, that one, okay, one sec.

jimthompson5910 (jim_thompson5910):

ok

OpenStudy (anonymous):

http://www.sketchtoy.com/48806079

OpenStudy (anonymous):

Yes?

jimthompson5910 (jim_thompson5910):

hmm not sure what you wrote there

jimthompson5910 (jim_thompson5910):

but you should have something like this m = (y2-y1)/(x2-x1) m = ( -11- 13)/(-4-2) m = ???

OpenStudy (anonymous):

Yeah, thats what I had written, I just have to do the last part.

jimthompson5910 (jim_thompson5910):

i gotcha, so what do you get for the slope

OpenStudy (anonymous):

-24/-6

jimthompson5910 (jim_thompson5910):

keep going

OpenStudy (anonymous):

Simplify it?

jimthompson5910 (jim_thompson5910):

yes reduce as much as possible

OpenStudy (anonymous):

-12/-3

jimthompson5910 (jim_thompson5910):

keep going

OpenStudy (anonymous):

I don't think I can simplify any more on the -3, can I?

jimthompson5910 (jim_thompson5910):

ignore the negatives for a second

jimthompson5910 (jim_thompson5910):

12/3 = ??

OpenStudy (anonymous):

4?

jimthompson5910 (jim_thompson5910):

so -12/(-3) = 4 as well because the two negatives divide to a positive

jimthompson5910 (jim_thompson5910):

this means that the slope of the line through (2, 13) and (−4, −11) is m = 4

jimthompson5910 (jim_thompson5910):

any line parallel to this will have an equal slope

OpenStudy (anonymous):

Thanks Jim! :)

jimthompson5910 (jim_thompson5910):

you're welcome

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