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Math Help - Find the equation of a line perpendicular to a line with a given equation

  1. #1
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    Find the equation of a line perpendicular to a line with a given equation



    Find the equation perpendicular to 10x 9y + 2 = 0 and passing through point (3,3) is given by...


    This is the answer given:


    Eqn. of reqd line is of form 9x+10y+k = 0; k is to be determined.
    (3,3)belongs to line=> 27+30+k = 0 or k = -57
    9x+10y-57 = 0
    Ans: C


    I set up the point-slope :


    y=10/9x + 2/9


    m for the perp line is -9/10, but then how do I solve for the intercept?


    y = -9/10x +c


    but then I set


    10y=-9x+10c



    10y+9x=10c


    c= y + 9/10x - that isn't right!
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by fcabanski View Post


    Find the equation perpendicular to 10x 9y + 2 = 0 and passing through point (3,3) is given by...


    This is the answer given:


    Eqn. of reqd line is of form 9x+10y+k = 0; k is to be determined.
    (3,3)belongs to line=> 27+30+k = 0 or k = -57
    9x+10y-57 = 0
    Ans: C


    I set up the point-slope :


    y=10/9x + 2/9


    m for the perp line is -9/10, but then how do I solve for the intercept?


    y = -9/10x +c


    but then I set


    10y=-9x+10c



    10y+9x=10c


    c= y + 9/10x - that isn't right!
    Hi fcabanski.

    What you are doing is correct. The k to be determined is equal to -10c.
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    So how do I arrive at the answer as was given in the solution?
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    I just told you. Your c is equal to \frac{57}{10} (set (x,y)=(3,3) in c=y+\frac9{10}x); hence k=-10c.
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  5. #5
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    That makes no sense. Normally I don't have to set k=nc.

    For example in this problem:

    Write an equation of the line
    perpendicular to 4y - x = 20 and
    containing the point (2, -3).


    y = 1/4x +20

    slope perpendicular line = -4

    y = -4x +b

    Plug in x and y

    -3 = -4(2) + b

    -3 = -8 +b

    5=b

    Perp line is therefore

    y = -4x +5

    Or

    4x + y - 5 = 0

    there's no setting an arbitrary k = nc


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  6. #6
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by fcabanski View Post
    That makes no sense.
    What isn’t making sense? You just have not understood it yet (though I thought you were so close to doing so).
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    Please explain it. In the second example I showed, there was no setting k=nc.

    Why is K=-10c set? How do you know to do it before you have the solution?
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  8. #8
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by fcabanski View Post
    Please explain it. In the second example I showed, there was no setting k=nc.

    Why is K=-10c set? How do you know to do it before you have the solution?
    Forget your second example. It was just a red herring.

    Let me put it this way. The equation of the line can be written as either

    10y+9x-57\ =\ 0

    or

    y\ =\ -\frac9{10}x+\frac{57}{10}

    They are the same. Yes? If you write it the first way, the constant term appears as -57. If the second way, it appears as \frac{57}{10}. Does this now make sense?
    Last edited by TheAbstractionist; June 2nd 2009 at 06:53 PM.
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  9. #9
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    Yes, but how do I get there from the slope point form?

    Forget we have the answer.

    Find the equation perpendicular to 10x – 9y + 2 = 0 and passing through point (3,3) is given by...

    -9y =-10x -2

    y = 10/9x +2/9

    Now we have the slope of the perp line = -9/10

    Perp Line y = -9/10x + b

    So we can find the y int given the point (3,3)

    3 = -9/30 +b

    What, without having the solution, would tell us to set k = -10 b? (I don't even know where I got c )

    We'd just solve:

    b = 3 + 9/30




    So that's what I'm asking, without that solution, how do we know to set some arbitrary k?
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  10. #10
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    m for the perp line is -9/10, but then how do I solve for the intercept?
    y - y1 = m (x - x1)

    y - 3 = -9/10 (x - 3)

    y - 3 = -9x/10 + 27/10

    y = -9x/10 + 57/10


    When you know a point on a line and the slope of the line, just grind them into point-slope form for lines and get your line. No reason to get all wordy.
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  11. #11
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by fcabanski View Post
    So we can find the y int given the point (3,3)

    3 = -9/30 +b
    You made a typo this time. It should be 3\ =\ -\frac{27}{10}+b.
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  12. #12
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    OK, so

    3 = -27/10+ b

    b = 3 +27/10 = 57/10

    so


    y = -9/10x +57/10

    10y = -9x +57

    9x + 10y -57 = 0


    I was forgetting to multiply the 57/10 by 10 to get the final equation.

    Now I see the mistake I was making since yesterday. Thanks.
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