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Straight Lines

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data

    There is a point A(-5,-4) through which th line L1 passes.
    Straight lines
    1)x+3y+2=0
    2)2x+y+4=0
    3)x-y-5=0
    intersect L1 at B,C,D respectively! It is known that:

    [tex]\frac{15^{2}}{AB^{2}}+\frac{10^{2}}{AC^{2}}=\frac{6^{2}}{AD^{2}}[/tex]

    Find equation of Line L1.
    2. Relevant equations
    General equation of straight line
    y=mx+c

    Distance(r) of a point(x,y) from A along the line L1
    [tex]\frac{x+5}{cos \theta}=\frac{y+4}{sin \theta}=r[/tex]

    where theta is the angle made with positive x axis by L1
    3. The attempt at a solution

    Let L1: y=mx+c
    (-5,-4) lies on the line
    therefore:
    L1: y=mx+4m-5
    where m is slope

    I solved this equation simultaneously with the three given above to get coordinates of intersecting points in terms of 'm'.
    I used the distance along line formula to get AB,AC,AD in terms of m. I plug it back to
    [tex]\frac{15^{2}}{AB^{2}}+\frac{10^{2}}{AC^{2}}=\frac{6^{2}}{AD^{2}}[/tex]

    what I get is the following beautiful equation in 'm'

    [tex]\frac{225(3m+1)^{2}}{(2m-5)^2}+\frac{100(1-m)^{2}}{(9m-5)^2}=\frac{9(m+2)^{2}}{(2m-5)^2}[/tex]

    First of all, I cannot solve for 'm'! Second thing is that the method I adopted was a tedious one. Someone please help me with a more clever and short technique.
     
  2. jcsd
  3. Nov 15, 2008 #2
    Are my equations in m correct?

    Just give it a thought to my second method. Will converting the coordinates of the intersecting point to parametric forms an then applying the distance along a line from a point formula help??

    Distance(r) of a point(x,y) from A along the line L1
    [tex]\frac{x+5}{cos \theta}=\frac{y+4}{sin \theta}=r[/tex]

    where theta is the angle made with positive x axis by L1
     
  4. Nov 16, 2008 #3

    Mark44

    Staff: Mentor

    I replied before I understood what you were trying to do. Let me think about it further.
     
  5. Nov 16, 2008 #4

    Mark44

    Staff: Mentor

    Your equation for L1 is incorrect. It's given that A(-5, -4) is on line L1, so its equation is y - (-4) = m(x - (-5), or
    y + 4 = mx + 5m, or
    y = mx + 5m - 4

    You have
     
  6. Nov 17, 2008 #5
    Oh, I am sorry. My calculations were wrong. but still I would get a nasty equation in m. Wouldnt I?
    Isnt there any simple method?
     
  7. Nov 17, 2008 #6

    Mark44

    Staff: Mentor

    I don't see any simpler approach.
     
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