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Thread: how to find the ratio

  1. #1
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    how to find the ratio

    Question: Given that ax+ 3y + 2 = 0 and 2x - by - 5 = 0 are perpendicular lines, find the ratio a:b.


    My attempt:
    $ax + 3y + 2 = 0 $
    $y = -\frac{2}{3} - \frac{a}{3}x$
    $2x - by - 5= 0 $
    $y = \frac{2}{b}x - \frac{5}{b}$
    $ m X - \frac{1}{m} =-1 $
    therefore

    $-\frac{a}{3} = -\frac{2}{b}.......(1)$

    $ m = -\frac{a}{3}$

    thus $ -\frac{1}{m} = -\frac{1}{-\frac{a}{3}} = \frac{3}{a}$
    $\frac{3}{a}= \frac{2}{b}......(2)$

    using simultaneous equation for 1 and 2
    you get $a = \pm 3$
    and

    $b = \pm 2$

    therefore the ratio i got is 3:2 or -3:-2


    the problem is that the book only states 3:2

    in the case of this question i check both positive and negative and my answrs are right but the book only states 3:2 is that just a mistake in the book?
    or did my method allowed an extraneous result?
    Last edited by bigmansouf; Feb 19th 2017 at 08:32 PM.
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  2. #2
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    Re: how to find the ratio

    Quote Originally Posted by bigmansouf View Post
    Question: Given that ax+ 3y + 2 = 0 and 2x - by - 5 = 0 are perpendicular lines, find the ratio a:b
    If neither $m_1~\&~m_2$ is zero and are the slopes of perpendicular lines $\ell_1~\&~\ell_2$ then $m_1\cdot m_2=-1$.

    So in your case $\dfrac{-a}{3}\cdot\dfrac{-2}{-b}=-1$
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  3. #3
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    Re: how to find the ratio

    Quote Originally Posted by bigmansouf View Post
    Question: Given that ax+ 3y + 2 = 0 and 2x - by - 5 = 0 are perpendicular lines, find the ratio a:b.


    My attempt:
    $ax + 3y + 2 = 0 $
    $y = -\frac{2}{3} - \frac{a}{3}x$
    $2x - by - 5= 0 $
    $y = \frac{2}{b}x - \frac{5}{b}$
    $ m X - \frac{1}{m} =-1 $
    therefore

    $-\frac{a}{3} = -\frac{2}{b}.......(1)$

    $ m = -\frac{a}{3}$

    thus $ -\frac{1}{m} = -\frac{1}{-\frac{a}{3}} = \frac{3}{a}$
    $\frac{3}{a}= \frac{2}{b}......(2)$

    using simultaneous equation for 1 and 2
    you get $a = \pm 3$
    and

    $b = \pm 2$

    therefore the ratio i got is 3:2 or -3:-2


    the problem is that the book only states 3:2

    in the case of this question i check both positive and negative and my answrs are right but the book only states 3:2 is that just a mistake in the book?
    or did my method allowed an extraneous result?
    -3:-2 =3:2

    a=6 and b=4 also satisfies your equations.
    6:4 = 3:2. You are asked for the ratio a:b, not a and b separately.
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    Re: how to find the ratio

    Quote Originally Posted by Plato View Post
    If neither $m_1~\&~m_2$ is zero and are the slopes of perpendicular lines $\ell_1~\&~\ell_2$ then $m_1\cdot m_2=-1$.

    So in your case $\dfrac{-a}{3}\cdot\dfrac{-2}{-b}=-1$
    i understand this but how do you get the ratio from this.

    2a=3b
    but how does it become 3:2 as a ratio thats my problem
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    Re: how to find the ratio

    Quote Originally Posted by bigmansouf View Post
    i understand this but how do you get the ratio from this.

    2a=3b
    but how does it become 3:2 as a ratio thats my problem
    2a=3b

    a=\frac{3b}{2}

    \frac{a}{b}=\frac{3}{2}

    a:b = 3:2
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  6. #6
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    Re: how to find the ratio

    Quote Originally Posted by bigmansouf View Post
    i understand this but how do you get the ratio from this.
    2a=3b but how does it become 3:2 as a ratio thats my problem
    Oh come on! Are you really saying that if $2a=3b$ that you fail to understand that $\dfrac{a}{b}=\dfrac{3}{2}~\bf{??}$
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  7. #7
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    Re: how to find the ratio

    Perhaps bigmansouf is saying that he does not know what the word "ratio" means.
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  8. #8
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    Re: how to find the ratio

    Quote Originally Posted by Plato View Post
    Oh come on! Are you really saying that if $2a=3b$ that you fail to understand that $\dfrac{a}{b}=\dfrac{3}{2}~\bf{??}$
    i understood 2a = 3b but not how to get it to a ratio form. I understood the relationship which you get from m1 x m2 = -1 but from there i failed to rearrange it to give a/b = 3/2 instead i thought of using simultaneous equation to help me find the ratio hence when i did that as shown in my first post i thought i got an extra answer which i believed to be extraneous result.

    The problem was i thought the question was far more complex than it was. i simply failed to spot it
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