since the lines are not parrallel, there is a common point for both lines,
take the point as the common point for both and .
now try to prove that
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?
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?
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 ?