Question

If a, b, c are in A.P., prove that the straight lines $ax+2y+1=0,bx+3y+1=0$ and $cx+4y+1=0$ are concurrent.

Answer 1

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

$b-a=c-b$

$2b=a+c$ [Common difference]

To prove that the straight lines are concurrent then they have the common point of intersection.

$ax+2y+1=0...\left(1\right)$

$bx+3y+1=0...\left(2\right)$

$cx+4y+1=0...\left(3\right)$

Solving (1) and (2)

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

Put in (2)

$b\left(\frac{-1-2y}{a}\right)+3y+1=0$

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

$=\frac{-3a+2b-2b+2a}{a(3a-2b)}$

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

Putting x, y in (3)

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

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

$-a+2b-c=0$

$-a+a+c-c=0$

$0=0$

Hence, proved.