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Math Help - Cordinate geometry

  1. #1
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    Coordinate geometry

    Dear forum members,

    could you please check if my calculations are correct, I have tried several times, I alway keep getting the wrong answer though.



    Problem

    Find the values for k for which the lines 2x+ky-11=0 and (3-k)x+y+2=0  are perpendicular.



    My solution

    -3x+kx= \frac{k}{2x}
    -3x+kx and 2x have to be the same

    so -3x+kx= 2x


    I always keep getting 5 or 1 as the answer for k, is it correct?








    Thank you in advance!
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  2. #2
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    Quote Originally Posted by Coach View Post
    Dear forum memebers,

    could you please check if my calculations are correct, I have tried several times, I alway keep getting the wrong answer though.



    Problem

    Find the values for k for which the lines 2x+ky-11=0 and (3-k)x+y+2=0  are perpendicular.

    My solution

    -3x+kx= //frac{k}{2x}
    the bottom numbers have to be the same

    so -3x+kx= 2x

    I always keep getting 5 or 1 as the anwer for k, is it correct?

    Thank you in advance!
    Two lines are perpendicular if the product of their gradients is equal to -1.

    2x+ky-11=0 \Rightarrow y = \frac{-2x+11}{k} \Rightarrow m = -\frac{2}{k}.

    (3-k)x+y+2=0 \Rightarrow y = -(3 - k)x - 2 = (k - 3)x - 2 \Rightarrow m = k - 3.

    Therefore you require \left( -\frac{2}{k} \right) \, (k-3) = -1 \Rightarrow -2(k-3) = -k \Rightarrow k = 6.
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  3. #3
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    Thank you so much!



    I would not have been able to do that on my own.
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  4. #4
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    Quote Originally Posted by Coach View Post
    Dear forum members,

    could you please check if my calculations are correct, I have tried several times, I alway keep getting the wrong answer though.

    Problem

    Find the values for k for which the lines 2x+ky-11=0 and (3-k)x+y+2=0  are perpendicular.

    I always keep getting 5 or 1 as the answer for k, is it correct?

    Thank you in advance!
    I just realised that if you kept getting 5 plus 1 as the answer for k it would be correct
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  5. #5
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    Quote Originally Posted by Coach View Post
    Thank you so much!



    I would not have been able to do that on my own.
    I'm sure you would You've just got to work through each link in the chain.
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