Question

If the lines $ax+2y+1=0$, $bx+3y+1=0$,$cx+4y+1-0$ are concurrent find relation between $a,b,c$

Answer 1

Given lines are

$ax+2y+1=0$----(1)

$bx+3y+1=0$----(2)

and $cx+4y+1=0$----(3)

solving(1) and (3)by cross-multiplication method. we get

$\frac{x}{2-3}=\frac{-y}{a-b}=\frac{1}{3a-2b}$

so,the point of intersection is $[\frac{1}{2b-3a},\frac{1}{3a-2b}]$

since,the point lies on cx+4y+1=0, then

$\frac{c}{2b-3a}+\frac{4(a-b)}{2b-3a}+1=0$

$\Rightarrow c+4a-4a+2b-3a=0$

$\Rightarrow 2b=a+c$

Hence a,b,c,are in A,P.