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Find slope of 2 perpendicular lines

  1. Jan 15, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the values of k such that lines 3kx+8y = 5 and 6y -4kx = -1 are perpendicular.

    I don't need the answer, but a push in a right direction. However, feel free to solve the equation :)

    2. Relevant equations
    1. Slope of a line is m = [y2-y1]/[x2-x1] where we have 2 points on the line
    P1(x1,y1) and P2(x2,y2)

    2. Product of slope of 2 perpendicular lines is -1
    let's say m1 is slope of line 1 and m2 is slope of line 2 then
    m1 = -(1/m2)

    3. The attempt at a solution

    My attempts to resolve this equation are just confusing. I know that we have too many variables and we need to get rid of some of them.

    I also know that at one point both equations are equal to each other (intersection point)

    3kx+8y = 5 and 6y -4kx = -1 are...

    1. k = (5-8y)/3x and 2. k = (-1 -6y)/ -4x

    if k = -(1/k) then

    (5-8y)/3x = -((-4x)/(-1-6y))

    (5-8y)(-1-6y) =(4x)*(3x)

    -5 -30y +8y +48y^2 = 12x^2

    and here I get stuck! What to do then? and am I taking a right path to solve this problem?

    Thank you very much for any help!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 15, 2007 #2


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    I think you're making it more difficult than it needs to be!

    You are correct that, if two lines are perpendicular then the product of their gradients is equal to -1. Try writing both equations in the form y=mx+c, where m is the gradient of the curve. Then you can envoke your relation m1m2=-1 and solve for k.
  4. Jan 15, 2007 #3
    thank you for your reply.

    It seems to me no matter how much I play with these 2 equations I always have either too many variables or get stuck with x and y square.

    I forgot to mention that k of line 1 isn't necessarily equals to k of line 2.
    So I don't see how writting them both in for y =mx +c will help me solve it.

    1. y = (5-3kx)/8 2. y =(-1+4kx)/6

    I'm still stuck with x being unknown and k of line 1 and k of line 2.
  5. Jan 15, 2007 #4


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    Ok, since the k's are different, write j in the first, and k in the second equation. Plug these into the formula m1m2=-1 and you will arrive at a relation between j and k.
  6. Jan 15, 2007 #5


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    cristo, I would not interpret this as two different values for k. The problem is asking for a value for k, in both equations, such that the two lines are perpendicular

    A more "relevant" equation is this: if y= mx+ b, then the slope is m. Can you solve 3kx+ 8y= 5 and 6y- 4kx= -1 so you have each in that form and can write the slope as a function of k only? Then use the fact that the product of the slopes must be -1.

    Last edited by a moderator: Jan 15, 2007
  7. Jan 15, 2007 #6


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    That's what I thought originally, but when the OP said:
    I thought he had been explicitly told otherwise!
  8. Jan 15, 2007 #7
    3kx+8y = 5 and 6y -4kx = -1 are perpendicular.
    slope of the first one can be obtained from
    y = 5/8 - 3k/8x
    and slope of the second one can be obtained from
    y = -1/6 + (2/3k)x

    from the first one the slope is -3k/8
    and from the second one 2k/3

    since the product of slopes is -1 you have

    -6k^2/24 = -1

    k^2/4 = 1

    k^2 = 4

    Check this, maybe I've messed something up, but according to this k is 2 and -2.
  9. Jan 15, 2007 #8
    y = mx + c can be used if they are the same. Are you trying to say that you need to find an equation for two different 'k's? e.g. k1 + k2 = c, where c is some constant. More like a linear programming question.

    You do not need to know x and at no stage do you need to. You are writing the equations in the form y = mx + c because you want to know what m is. x makes no difference to the gradient. Knowing the two 'm's from the two, you can use m1 x m2 = -1, as you stated. However........

    I concour with *best&sweetest*, although I do need to remind you that the person that asked the question is suppose to work through the steps and understanding to produce the solution.


    The Bob (2004 ©)
    Last edited: Jan 15, 2007
  10. Jan 15, 2007 #9
    As the solution for k1 = k2 has been given, I am going to go through a more general solution, which is more or less the same as post #7. All you have to do is replace the 'k's, one with k1 and the other with k2. This will produce k1 x k2 = 4, which is obvious from the posts before. So find any two values that satisfy this equation and you will have all of the solutions for k, assuming they differ.

    The Bob (2004 ©)
  11. Jan 15, 2007 #10


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    Are we all ignoring the homework help rules today? Please do not post full solutions. Instead try to help the OP to get the answer for himself!
  12. Jan 15, 2007 #11
    Whoaa you guys are the best!

    First of all, I'm terribly sorry for confusing you all with
    "I forgot to mention that k of line 1 isn't necessarily equals to k of line 2."

    This is my mind playing games on me after 2 days of thinking about this problem. Nowhere has it said that these 2 variables are not equal! For some reason I got an idea that they are and it lead me to a wrong direction.

    Thank you very much for all your input, it was very interesting to read your ideas and see the approaches you took to solve this problem, and the other problem that I introduced unintentionally hehe :-)

    best&sweetest thank you for solving the problem, I followed your example and verified that - (3k/8)*(2k/3) = -1 if K is +/- 2.

    HallsofIvy, I had tried your suggestion before posting this topic, but for some reason I thought it wasn't the right way and abandoned it in favor of more complex and confused path. Big Mistake!

    Bob and Cristo, thank you for your help!
  13. Jan 15, 2007 #12
    No, I have not. The answer for k1 = k2 had already been given. It was, even in words, pretty intuitive that the general solution was k1 x k2 = -1. All that could be have been said was instead of using k2 to replace it with two 'k's.

    I apologise for finishing the question.

    The Bob (2004 ©)
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