Question

If the lines $(a-b-c)x+2ay+2a=0$, $2bx+(b-c-a)y+2b=0$
and $(2c+1)x+2cy+(c-a-b)=0$ are concurrent, then prove that either
$a+b+c$ or $(a+b+c{)}^2+2a=0$

Answer 1

$(a-b-c)x+2ay+2a=0$

$2bx+(b-c-a)y+2b=0$

$(2c+1)x+2cy+(c-a-b)=0$

Since lines are concurrent, determinant should be:

$\left|\begin{array}{ccc}(a-b-c)& 2a& 2a\\ 2b& (b-c-a)& 2b\\ (2c+1)& 2c& (c-a-b)\end{array}\right|=0$

$=(a-b-c)\left[\right(b-c-a\left)\right(c-a-b)-(2b\left)\right(2c\left)\right]-2a\left[2b\right(c-a-b)-2b(2c+1\left)\right]+2a\left[\right(2b\left)\right(2c)-(b-c-a\left)\right(2c+1\left)\right]$

$=(a-b-c)[bc-\text{/}ab-b^2-c^2+\text{/}ac+bc-\text{/}ac+a^2+\text{/}ab-4bc]-2a[2bc-2ab-2b^2-4bc-2b]+2a[4ab-2bc-b+2c^2+c+2ac+a]$

$=(a-b-c)[a^2-b^2-c^2-2bc]-2a[-2bc-2ab-2b^2-2b]+2a[2bc-b+c+a+2c^2+2ac]$

$=a^2+2a+b^2+2bc+2ab+c^2+2ac$

$=(a^2+b^2+c^2+2ab+2bc+2ac)+2a$

$(a+b+c{)}^2+2a=0$.