The lines below are perpendicular. If the slope of the green line is 1, what is the slope of the red line ? http://media.apexlearning.com/Images/200706/18/055cb2ff-62ad-426e-a3db-d0871c0ab884.gif m=

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The lines below are perpendicular. If the slope of the green line is 1, what is the slope of the red line ? http://media.apexlearning.com/Images/200706/18/055cb2ff-62ad-426e-a3db-d0871c0ab884.gif m=

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Other answers:

okay
okay next ..
try an attempt the problem.. either let m_1 = 1 or m_2 = 1 then divide.
m_1=1
|dw:1436358056467:dw| k now what do we need to do next?
? divide ?
yeah
whenever perpendicular, they are both the same.....
-1/-1 ?
just flip the sign
the answer would be 1 ?
@waleedo212 wrong. parallel lines have the same slope
oh okay nvm
@waleedo212 perpendicular lines have different slopes.
\[m_2 = \frac{-1}{1} \]
so divide -1/-1 ?
*points up*
okay
its 1
no!
oh can you help me out on that one .
scroll up to what I've Latex'd
m1xm2=-1
\[m_2 = \frac{-1}{1} \]
okay what would that be ? 1
... -1 divided by 1 = ? 1 divided by 1 = 1 so NEGATIVE 1 divided by 1 is? Anything divided by 1 is the number on top of the fraction
-1
yess
thats the answer then right ?
yes it is. you can always plug it back into the equation \[m_1 \times m_2 = -1 \rightarrow 1 \times -1 = -1 \]

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