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Math Help - strange line question

  1. #1
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    strange line question

    If the lines (A1)X + (B1)Y + (c1) = 0 and (A2)X + (B2)Y + (C2) = 0 are not parallel and K is any constant, show that :

    ((A1)X + (B1)Y + (c1)) + K((A2)X + (B2)Y + (C2)) = 0

    Is a line throught the point of intersection of the given lines. When K is assigned various values, this equaion represents various members of the family of all lines through the point of intersection.

    Can anyone give me any tips to solving this one ? I tried alot last night but couldnt figure it out. How Do I show that its true anyway? Plus you cant even add them together since there in parenthesies right? What should I do?
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  2. #2
    Senior Member BAdhi's Avatar
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    since the lines are not parrallel, there is a common point for both lines,

    take the point (x_0,y_0) as the common point for both a_1x+b_1y+c_1=0 and a_2x+b_2y+c_2=0.

    now try to prove that a_1x_0+b_1y_0+c_1+K(a_2x_0+b_2y_0+c_2)=0
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  3. #3
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    ok... but it doesnt really help me. I mean its obvious that it will equal 0 since a1x+b1y+c1 = 0 and a2x+b2y+c2=0 then its basicly saying 0 + K (0) = 0. so ofcoarse its 0. but how does it become a line? I also get that each 2 equations seperatly are lines. I mean, for the sake of doing anything all I can think to do is actually this :

    2y= (-a1x-c1-ka2x-kc2) / b1+kb2

    nothing there cancels out. putting x and y are x0,y0 doesnt make my brain think of anything new. what should I do?
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  4. #4
    Senior Member BAdhi's Avatar
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    a_1x+b_1y+c_1+K(a_2x+b_2y+c_2)=0

    this can be arranged as,
    (a_1+Ka_2)x+(b_1+Kb_2)y+(c_1+Kc_2)=0

    since a_1,a_2,b_1,b_2,c_1,c_2 and K are constants the above equation is similar to

    ax+by+c=0(equation of a line) which represents a line
    Last edited by BAdhi; February 15th 2011 at 06:06 PM.
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  5. #5
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    according to the question, (a_1+Ka_2)x+(b_1+Kb_2)y+(c_1+Kc_2)=0 is supposed to be a line that goes threw the point of intersection of a_1x+b_1y+c_1=0 and a_2x+b_2y+c_2=0 .

    So I take 2 lines : x+3y-2=0 and 2x-y+4=0.

    there point of intersection is (-2,0).

    now (a_1+Ka_2)x+(b_1+Kb_2)y+(c_1+Kc_2)=0 when applied to these 2 lines should find other lines with a similar point of intersection depending on K. these 2 lines threw this equation turns out as x(1+k2)+y(3+-1k)+(-2+k4)=0.

    Im getting confused though, because when I assign various values to K, im not getting a line with (-2,0) as the point of intersection. Why not ?
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  6. #6
    Senior Member BAdhi's Avatar
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    because (-2,0) is not your common point (substitute your values to the first line)
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