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

1. calculusxy

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

2. calculusxy

Am I correct?

3. calculusxy

@Michele_Laino

4. 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

5. 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: $\large {m_1}{m_2} > 0$

6. Michele_Laino

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

7. calculusxy

8. calculusxy

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

9. Michele_Laino

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

10. calculusxy

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

11. Michele_Laino

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

12. calculusxy

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

13. calculusxy

Is your and my equations the same thing? @Michele_Laino

14. Michele_Laino

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

15. calculusxy

16. calculusxy

Would that line work?

17. calculusxy

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

18. Michele_Laino

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

19. Michele_Laino

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

20. calculusxy

@Michele_Laino How did you get 13/5?

21. Michele_Laino

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

22. calculusxy

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

23. calculusxy

I am really sorry fir being redundant. :(

24. calculusxy

*for

25. Michele_Laino

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

26. calculusxy

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

27. Michele_Laino

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

28. Michele_Laino

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

29. calculusxy

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

30. Michele_Laino

:)

31. calculusxy

So for x I can input any number now?