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What is the proof of the formula L1 +pL2=0?

  1. May 1, 2012 #1
    1. The problem statement, all variables and given/known data
    This related to coordinate geometry.
    According to my book it says if there are two intersecting lines L1(a1x+b1y+c1=0) and L2(a2x+b2y+c2=0)
    Then equation of any line passing through their point of intersection is
    L1 +p L2 =0
    or (a1x+b1y+c1=0) +p(a2x+b2y+c2=0)=0
    where p is some constant
    my problem is that i don't understand what is the proof of this formula.Book also says that (a1x+b1y+c1=0) +p(a2x+b2y+c2=0)=0 represents a Family of lines , and i don't understand what it means too.

    2. Relevant equations



    3. The attempt at a solution
    i tried to use some arbitrary line equations and tried to simultaneously solve the equations
    and get its point of intersection and find the line's equation but i can't relate these 2 methods
    If k is such that it a1-p.a2=0 how does this proove these lines are concurrent and it equation is (a1x+b1y+c1=0) +p(a2x+b2y+c2=0)=0
    book gives the proof as since this line satisfies some P(a,b) hence when we put it into the equation it will yield 0
    But that comes only after i have proved this is the equation of all lines passing through common point ,how can this be a proof?

    Could you please give me a proof of how this formula is derived?
    Thank you.
     
    Last edited: May 1, 2012
  2. jcsd
  3. May 1, 2012 #2
    You have a line Lp defined by, (a1x+b1y+c1) +p(a2x+b2y+c2)=0. Say that lines L1, L2 intersect at a point P, then P is both on L1 and on L2, thus a1*Px+b1*Py+c1 = 0 and a2*Px+b2*Py+c2 = 0. 0+p*0 = 0 so P is also on Lp. That's all there is to it.

    A family of lines means that Lp is not any specific line. It could represent many different lines, if different values of p are used.
     
  4. May 1, 2012 #3
    I already know if P satisfies L1 and L2 it is on line L1 and L2 because basically equation of line L1 and L2 is a condition & when we graph it we get a line.But what i am asking is HOW DO WE KNOW EQUATION OF LINES PASSING THROUGH A POINT CAN BE GIVEN BY (a1x+b1y+c1=0) +p(a2x+b2y+c2=0)=0
    I WANT THE PROOF .

    One more thing i found was if we add or subtract the equations of two intersecting lines the resultant line will be concurrent to the other two lines
    And if it is correct(which i hope it is(!):redface:) then how can find if they are concurrent from their equations when the slope and the y-intercept always remains the same no matter which point on line we take.
     
  5. May 1, 2012 #4
    In the line equation ax + by + c = 0, (a, b) is a normal of the line. Another way to write the same equation is <(a, b), (x, y)> = -c, where < , > is the scalar product. It follows from properties of < , > that if <p, q> = w then if you take any vector n, orthogonal to p, <p, q+n> is also = w.

    Or was that not what you were asking again?
    What I gave you is a valid proof that if L1 and L2 both contain P, then so does L1+p*L2. It is not necessary to derive a thing to prove it.
     
  6. May 1, 2012 #5
    I don't understand!
     
  7. May 1, 2012 #6
    You'll have to be more specific. Do you know what a normal is? A scalar product? That <a, b> = 0 if a and b are perpendicular? That <a+b, c> = (ax+bx)*cx + (ay+by)*cy = ax*cx + ay*cy + bx*cx + by*cy = <a, c> + <b, c>?
     
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