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Intersection of Two Lines

  1. Jan 25, 2006 #1
    I have tried but doesn't work out well....

    Find the point at which the normal through the point (3,-4) to the line 10x+4y-101=0 intersects the line.

    Originaly I thought that it should be found by doing the dot product of:

    (x-3,y+4)dot(-4/10,-10/4)= 0
  2. jcsd
  3. Jan 25, 2006 #2


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    I can't help but wonder what your reasoning is! (The vector (-4/10, -10/4) is not in the direction of that line. The vector (4, -10) or (-4, 10) are.)

    The line 10x+ 4y- 101= 0 can be written as y= (-10/4)x+ 101/10. That has slope -10/4 so a normal line has slope 4/10= 2/5.
    The line through (3,-4) with that slope can be written y= (2/5)(x-3)- 4 or, without fractions, 4y= 2x- 26

    Solve the equations 10x+ 4y- 101= 0 and 4y= 2x- 26 for x and y.
  4. Jan 25, 2006 #3
    ok i don't get how you get y= (2/5)(x-3)- 4

    can't you do (x-3,y+4)dot(2,5)= 0 and then substitute

    can't I also write it as r=(3,-4)+t(2,5)?????

    LOL i tried doing it my way and i got as an answer to be (187/14,-171/21)
    Last edited: Jan 25, 2006
  5. Jan 25, 2006 #4
    lol can someone help me further :(
  6. Jan 26, 2006 #5


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    Are you determined to do this using vectors? Since this problem is entirely in 2 dimensions, I see no reason not to do it in the simplest possible way. The line 10x+4y-101=0 has slope -5/2. (Solve for y:
    4y= -10x+ 101 so y= (-10/4)x+ 101/4 and -10/4= -5/2. Any line perpendicular to that must have slope 2/5.

    The "point slope" form for a line, with slope m going through (a, b), is
    y= m(x-a)+ b. That clearly has slope m and, just as clearly, when x= a, y= b.
  7. Jan 26, 2006 #6
    yea but i'm doing geometry and you was is like going back to grade 10.....We learned vectors and probably have to do it taht way
  8. Jan 26, 2006 #7
    yea i''m really struggling with this one....

    I understand that if you change 10x+4y-101=0 to y=(101-10x)\4 that the direction vector for that line would be (4,-10) am I right?

    then since they are perpendicular I would take the point (3,-4) and subract it to make the direction vector for the other line....so

    (x-3,x+4)dot(4,-10)=0 AM I RIGHT?

    P.S. I have to do it this way because I have to use vectors
  9. Jan 27, 2006 #8


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    Part of your problem is that you keep saying the direction vector. There are an infinite number of vectors pointing in the direction of a given line, all of different lengths.

    Yes, if x increases by 4 (say from 0 to 4), y decreases by 10 (from 101/4 to 61/4) so a vector pointing along the line is (4, -10). You could just as easily use (2, -5) or (-2, 5).

    Yes, the equation of the perpendicular line can be written
    (x- 3, y+ 4)dot(4, -10)= 0 (You have "(x- 3, x+ 4)" but I assume that is a typo. If you multiply that out, you get 4(x- 3)- 10(y+ 4)= 4x- 12- 10y- 40= 4x- 10y- 52=0 which is equivalent to the equation I gave.

    Solve the two equations, 4x- 10y- 52=0 and 10x+4y-101=0 simultaneously to find the point at which they intersect.
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