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Thread: Slopes of perpendicular lines..

  1. #1
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    Slopes of perpendicular lines..

    if

    a and 3
    -
    2

    are the slopes of perpendicular lines, a =

    someone wanna explain negative recipricals to me?
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by zoso
    if

    a and 3
    -
    2

    are the slopes of perpendicular lines, a =

    someone wanna explain negative recipricals to me?
    you got it a/2 *3=-1
    a=-2/3

    Malay
    Last edited by malaygoel; Jul 18th 2006 at 09:19 PM.
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  3. #3
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    Quote Originally Posted by zoso
    if

    a and 3
    -
    2

    are the slopes of perpendicular lines, a =

    someone wanna explain negative recipricals to me?
    What do you mean a=? Are you just using it as another variable to the slope (m)? Usually the slopes of all lines are m=, regardless of them being perpendicular. In your case, the slope of the line perpendicular to a line with the slope of$\displaystyle 3/2$ would be $\displaystyle m=-2/3$

    Negative reciprocals are basically just the opposite of the original fraction/whole number (just flip it).

    Examples:
    $\displaystyle m=3/1$, the negative reciprocal is: $\displaystyle -1/3$
    $\displaystyle m=5/2$, the negative reciprocal is: $\displaystyle -2/5$
    $\displaystyle m=-1/5$, the negative reciprocal is: $\displaystyle 5$
    $\displaystyle m=-4/5$, the negative reciprocal is: $\displaystyle 5/4$

    Remember, parallel lines have the same slope.

    Hope this helps..
    -NineZeroFive
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  4. #4
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by 905
    What do you mean a=? Are you just using it as another variable to the slope (m)? Usually the slopes of all lines are m=, regardless of them being perpendicular. In your case, the slope of the line perpendicular to a line with the slope of$\displaystyle 3/2$ would be $\displaystyle m=-2/3$

    Negative reciprocals are basically just the opposite of the original fraction/whole number (just flip it).

    Examples:
    $\displaystyle m=3/1$, the negative reciprocal is: $\displaystyle -1/3$
    $\displaystyle m=5/2$, the negative reciprocal is: $\displaystyle -2/5$
    $\displaystyle m=-1/5$, the negative reciprocal is: $\displaystyle 5$
    $\displaystyle m=-4/5$, the negative reciprocal is: $\displaystyle 5/4$

    Remember, parallel lines have the same slope.

    Hope this helps..
    -NineZeroFive
    the question was if a line with the slope $\displaystyle \frac{a}{2}$ is perpendicular to a line with the slope $\displaystyle 3$ then what is the value of a?

    which (this is just extending what Malay said) would mean that $\displaystyle 3$ and $\displaystyle \frac{a}{2}$ are negative reciprocal, which implies that multiplied together they equal -1, now solve for a...

    $\displaystyle \frac{a}{2}\times 3=\neg 1$

    $\displaystyle \frac{a}{2}=\frac{\neg 1}{3}$

    $\displaystyle \boxed{a=\frac{\neg 2}{3}}$

    check:

    $\displaystyle \frac{a}{2}\times 3=\neg 1$

    $\displaystyle \frac{\frac{\neg 2}{3}}{2}\times 3=\neg 1$

    $\displaystyle \frac{\neg 2}{6}\times 3=\neg 1$

    $\displaystyle \frac{\neg 6}{6}=\neg 1$

    $\displaystyle \neg 1=\neg 1$


    Interesting use of negative reciprocal:

    Just a bit of info you might like nzf, in a linear equation like $\displaystyle y=mx+b$ to find the x-intercept you multiply the y-intercept by the slopes negative reciprocal.

    ~ $\displaystyle Q\!u\!i\!c\!k$
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  5. #5
    Super Member malaygoel's Avatar
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    Slope are used to find family of lines.
    You are given a line $\displaystyle ax+by+c=0$

    Family of lines parallel to the given line
    $\displaystyle ax+by+k=0$(Same coefficients of x and t means same slope, k is a parameter)

    Family of lines perpendicular to the given line
    $\displaystyle bx-ay+k=0$
    or,$\displaystyle -bx+ay+k=0$(Swapping the cofficients of x and y and multiplying anyone of then by -1 gives a perpendicular line, this is due to negative reciprocal)

    Keep Smiling
    Malay
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