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

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calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0I think I got it. The slope needs to be negative reciprocals of the other. So like 5x needs to be 1/5x?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No. 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/coordinategeometry/slopesparallelandperpendicularlines

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4if 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
 one year ago
Best ResponseYou've already chosen the best response.4it is a contradiction, since we know that the product of such slopes has to check this condition: \[\large {m_1}{m_2} =  1\]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0So I need to make a perpendicular line through the 2,3 point. How can I do that?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4it is simple the slope of the requested line, is: \[\large m =  \frac{1}{5}\]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0I have the slope intercept form for 2,3 as y=5x + 2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4so you have to apply this equation: \[\large y  3 =  \frac{1}{5}\left( {x + 2} \right)\]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0So now do I need to do: 1/5x + 2= y?

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0Is your and my equations the same thing? @Michele_Laino

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4I 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
 one year ago
Best ResponseYou've already chosen the best response.0Would that line work?

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0My line doesn't work. It is not perpendicular.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4I think not, since that line is not perpendicular to both the parallel lines

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4please try with this line: \[\large y =  \frac{x}{5} + \frac{{13}}{5}\]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino How did you get 13/5?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4here 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0I am really sorry fir being redundant. :(

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4hint: \[ \Large  \frac{2}{5} + 3 = \frac{{  2 + \left( {3 \times 5} \right)}}{5} = ...?\]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0How did you get 2+(3 x 5)?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4since the least common multiple of 5 and 1 is 5 \[\Large  \frac{2}{5} + 3 =  \frac{2}{5} + \frac{3}{1}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4and 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay!!!!! I totally get it now :)

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0So for x I can input any number now?
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