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 find $a+b+c$

Answer 1

$(a-b-c)x+2ay+2a=02bx+(b-c-a)y+2b=0(2c+1)x+2cy+c-a-b=0$

The above 3 lines are concurrent

$\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c+1& 2c& c-a-b\end{array}\right|=0=>(a-b-c)\left[\right(b-c-a\left)\right(c-a-b)-4bc]-2a\left[2b\right(c-a-b)-2b(2c+1\left)\right]+2a[4bc-(2c+1\left)\right(b-c-a\left)\right]=0=>(a-b-c)[bc-c^2-ac-ab+ac+a^2-b^2+bc+ab-4bc]-2a[2bc-2ab-2b^2-4bc-2b]+2a[4bc-2bc-b+2c^2+c+2ac+a]=0=>(a-b-c)[a^2-b^2-c^2-2bc]+2a[2ab+2bc+2b+2b^2]+2a[2bc+a-b+c+2ac+2c^2]=0=>a^3-a^{2}b-a^{2}c-a{b}^2+b^3+b^{2}c-a{c}^2+b{c}^2+c^3-2abc+2b^{2}c+2b{c}^2+4a^{2}b+4abc+4ab+4a{b}^2+4abc+2a^2-2ab+2ac+4a^{2}c+4a{c}^2=0=>a+b+c=0$