calculusxy
  • calculusxy
MEDAL!!! Is it possible for two lines with positive slopes to be perpendicular to each other?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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calculusxy
  • 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
  • calculusxy
Am I correct?
calculusxy
  • calculusxy
@Michele_Laino

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anonymous
  • 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
  • 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
  • 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
  • calculusxy
calculusxy
  • calculusxy
So I need to make a perpendicular line through the -2,3 point. How can I do that?
Michele_Laino
  • Michele_Laino
it is simple the slope of the requested line, is: \[\large m = - \frac{1}{5}\]
calculusxy
  • calculusxy
I have the slope intercept form for -2,3 as y=5x + 2
Michele_Laino
  • Michele_Laino
so you have to apply this equation: \[\large y - 3 = - \frac{1}{5}\left( {x + 2} \right)\]
calculusxy
  • calculusxy
So now do I need to do: -1/5x + 2= y?
calculusxy
  • calculusxy
Is your and my equations the same thing? @Michele_Laino
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
  • calculusxy
calculusxy
  • calculusxy
Would that line work?
calculusxy
  • calculusxy
My line doesn't work. It is not perpendicular.
Michele_Laino
  • Michele_Laino
I think not, since that line is not perpendicular to both the parallel lines
Michele_Laino
  • Michele_Laino
please try with this line: \[\large y = - \frac{x}{5} + \frac{{13}}{5}\]
calculusxy
  • calculusxy
@Michele_Laino How did you get 13/5?
Michele_Laino
  • 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
  • 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
  • calculusxy
I am really sorry fir being redundant. :(
calculusxy
  • calculusxy
*for
Michele_Laino
  • Michele_Laino
hint: \[ \Large - \frac{2}{5} + 3 = \frac{{ - 2 + \left( {3 \times 5} \right)}}{5} = ...?\]
calculusxy
  • calculusxy
How did you get -2+(3 x 5)?
Michele_Laino
  • 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
  • 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
  • calculusxy
Oh okay!!!!! I totally get it now :)
Michele_Laino
  • Michele_Laino
:)
calculusxy
  • calculusxy
So for x I can input any number now?

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