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Determine point on line where normal passes thr a point intersects the line.

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine the point on the line 5x-4y-2=0 where the normal that passes through the point (7,-2) intersects the line.

    3. The attempt at a solution

    I'm completely and hopelessly lost. Any help or tips would be greatly appreciated. Thanks so much in advance!
     
  2. jcsd
  3. Mar 3, 2009 #2

    HallsofIvy

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    Presumably you know that the normal to y= mx+ b has slope -1/m.
    5x- 4y- 2= 0 is the same as 4y= 5x- 2 or y= (5/4)x- 1/2. Its slope is 5/4 so the slope of any normal to it is -4/5.

    Now, what is the equation of the line through (7, -2) with slope -4/5?
     
  4. Mar 4, 2009 #3
    Well a Vector Equation for that would be r = (7,-2) + t(-4,5) right? So where do I go from here?

    Am I supposed to find the parametric equations of each and find the point of intersection? If so, what steps do I take from the ones mentioned above?
     
  5. Mar 4, 2009 #4

    HallsofIvy

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    Since you are given one line in Cartesian, xy, form, it would be better to write the second line that way as well. I would be very surprised if you were working with vectors and parametric equations but did not know the "point, slope" form for a line in the plane.
     
  6. Mar 4, 2009 #5
    So for the first line I have:

    x = 7 - 4t
    y= -2 + 5t

    for the second line:

    x = (4y + 2)/5

    y = (5x - 2)/4


    Now normally I'd equate the two, and solve. However there is no 't' value in the 2nd line from what I gathered? Am I missing something?

    Sorry if I'm being a little frustrating, I had pneumonia and missed nearly all the lessons. :(
     
  7. Mar 4, 2009 #6
    After some calculations I've come to the possible answer of (-2,-3).

    Is this correct?
     
  8. Mar 4, 2009 #7
    Scratch that, I got (2,2)
     
  9. Mar 4, 2009 #8
    If you have a line with the equation of 5x-4y-2=0, you can figure out the slope of it's normal. If you have the slope and a point which the normal crosses (7; -2), all that you lack from normal equation y=kx+b is b. You can figure out slope from the other equation and a point with it's x and y coordinates is given. Sou you have system of linear equations to solve, where the solution is the crosspoint. I hope that helps.
     
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