MEDAL!!!
Is it possible for two lines with positive slopes to be perpendicular to each other?

- calculusxy

MEDAL!!!
Is it possible for two lines with positive slopes to be perpendicular to each other?

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- calculusxy

I think I got it. The slope needs to be negative reciprocals of the other. So like 5x needs to be -1/5x?

- calculusxy

Am I correct?

- calculusxy

@Michele_Laino

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## More answers

- anonymous

No. Theres no way for a line to cross another and form right angles unless one has a negative slope and one has a positive slope.
"If two lines are perpendicular and neither one is vertical, then one of the lines has a positive slope, and the other has a negative slope. Also, the absolute values of their slopes are reciprocals."
http://www.cliffsnotes.com/math/geometry/coordinate-geometry/slopes-parallel-and-perpendicular-lines

- Michele_Laino

if two perpendicular lines have their slopes , say m_1, and m_2 both positive, then the product of such slopes is also positive, namely:
\[\large {m_1}{m_2} > 0\]

- Michele_Laino

it is a contradiction, since we know that the product of such slopes has to check this condition:
\[\large {m_1}{m_2} = - 1\]

- calculusxy

##### 1 Attachment

- calculusxy

So I need to make a perpendicular line through the -2,3 point.
How can I do that?

- Michele_Laino

it is simple the slope of the requested line, is:
\[\large m = - \frac{1}{5}\]

- calculusxy

I have the slope intercept form for -2,3 as y=5x + 2

- Michele_Laino

so you have to apply this equation:
\[\large y - 3 = - \frac{1}{5}\left( {x + 2} \right)\]

- calculusxy

So now do I need to do: -1/5x + 2= y?

- calculusxy

Is your and my equations the same thing? @Michele_Laino

- Michele_Laino

I think that they are different equations, since after a simplification, I can rewrite my equation as follows:
\[\large y = - \frac{x}{5} + \frac{{13}}{5}\]

- calculusxy

##### 1 Attachment

- calculusxy

Would that line work?

- calculusxy

My line doesn't work. It is not perpendicular.

- Michele_Laino

I think not, since that line is not perpendicular to both the parallel lines

- Michele_Laino

please try with this line:
\[\large y = - \frac{x}{5} + \frac{{13}}{5}\]

- calculusxy

@Michele_Laino How did you get 13/5?

- Michele_Laino

here are the missing steps:
\[\large \begin{gathered}
y - 3 = - \frac{x}{5} - \frac{2}{5} \hfill \\
\hfill \\
y = - \frac{x}{5} - \frac{2}{5} + 3 \hfill \\
\hfill \\
y = - \frac{x}{5} + \frac{{15 - 2}}{5} \hfill \\
\hfill \\
y = - \frac{x}{5} + \frac{{13}}{5} \hfill \\
\end{gathered} \]

- calculusxy

I still do not get how you got 13/5.
These are my steps:
\[y - 3 = -\frac{ 1 }{ 5 } (x+2)\]
\[y - 3=-\frac{ 1 }{ 5 }x-\frac{ 2 }{ 5 }\]
\[y=-\frac{ 1 }{ 5 }x-\frac{ 2 }{ 5 } + 3\]
\[y=-\frac{ 1 }{ 5 }x+\frac{ 3 }{ 5 }\]

- calculusxy

I am really sorry fir being redundant. :(

- calculusxy

*for

- Michele_Laino

hint:
\[ \Large - \frac{2}{5} + 3 = \frac{{ - 2 + \left( {3 \times 5} \right)}}{5} = ...?\]

- calculusxy

How did you get -2+(3 x 5)?

- Michele_Laino

since the least common multiple of 5 and 1 is 5
\[\Large - \frac{2}{5} + 3 = - \frac{2}{5} + \frac{3}{1}\]

- Michele_Laino

and multipling both numerator and denominator of the second fraction, by 5, we get:
\[\large \frac{2}{5} + \frac{3}{1} = - \frac{2}{5} + \frac{{3 \times 5}}{{1 \times 5}} = - \frac{2}{5} + \frac{{15}}{5} = \frac{{ - 2 + 15}}{5}\]

- calculusxy

Oh okay!!!!! I totally get it now :)

- Michele_Laino

:)

- calculusxy

So for x I can input any number now?

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