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Simple Linear Algebra question

  1. Jan 11, 2015 #1
    1. The problem statement, all variables and given/known data

    A line is perpendicular to the line 2x - 4y + 7 = 0 and that passes through the point P(7,2). Determine the equation of this line in Cartesian form.

    2. Relevant equations

    3. The attempt at a solution
    Okay, so how I generally believe how to solve these problems may be wrong but i'm not sure.
    1.The normal vector is (2, -4) so by using this info i found the directional vector to be (4,2)
    2. After finding the directional vector, i used the dot product to find a vector that is perpendicular to (4,2) so i got (-2,4). I believe that this vector must be parallel to (2,-4) because otherwise it doesn't make sense(atleast to me)
    3. After finding the normal vector to (4,2), i used this information to simulate part of the Cartesian Equation.
    4. -2x + 4y = -C; What I did from here was use the specific point i was given (7,2) and implemented it in to the x and y values. I got -2(7) + 4(2) = -6
    5. My cartesian equation is essentially -2x + 4y -6 = 0.

    I'm getting the right answers for some questions but the wrong one for this one, can anyone help out?
  2. jcsd
  3. Jan 11, 2015 #2


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    Right there, you can use (4,2) as the normal vector to your perpendicular line so try ##4x+2y = C## and make it pass through your point.
  4. Jan 11, 2015 #3
    Okay i see, but it looks like my fundamental understanding of the problem is incorrect, or I simply don't understand the cartesian equation.

    Line 1's normal vector is -2,4
    Line 2's normal vector is 4,2
    Are these normal vectors collinear or perpendicular? I assumed they had to be collinear
    Because ((4)(2) + (-4)(2)) = 0; I'm confused about this.
  5. Jan 11, 2015 #4


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    If the lines are perpendicular their normal vectors are perpendicular. That's why their dot product is zero. Draw a picture.
  6. Jan 12, 2015 #5
    Yes that was what i was wondering; Thanks for your help.
    I assumed that they could be either perpendicular, or the normal vectors could be collinear as well.
  7. Jan 12, 2015 #6


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    You don't talk about vectors being collinear. You probably mean parallel. Points can be collinear, meaning in a straight line. And, of course, if the normal vectors were parallel, the lines would be parallel, not perpendicular.
  8. Jan 12, 2015 #7
    I know, i was unable to communicate it properly. I know that for any pair of lines, their angles between their normal vectors will be the same. Again, thanks for the help
  9. Jan 13, 2015 #8


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    Another way of looking at this- although it is no longer "Linear Algebra" just "Calculus":
    The line 2x- 4y= 7 can be rewritten as y= (1/2)x- 7/4. It has slope 1/2. A line perpendicular to it must have slope -2. So you want to find a line with slope -2, through point (7, 2). Perhaps you can use that to check your answer.
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