Question

If the lines $ax+y=0,$ $x+ay+2=0$ and $x+y+a=0(a\in R)$ are concurrent lines, then sum then sum of all possible value(s) of a is

• A. $0$
• B. $-1$
• C. $-2$
• D. $4$

Answer 1

The correct option is A $0$

$ax+y=0\to \left(1\right)$ $x+ay+2=0\to \left(2\right)$ $x+y+a=0\to \left(3\right)$

$a\times \left(3\right)-\left(2\right)$

$ax+ay+a^2=0$

$\underset{–}{x+ay+2=0}$

$(a-1)x=2-a^2$

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

Substituting $x$ in equation $\left(3\right)$

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

$=\frac{{-a}^2+a+a^2-2}{a-1}=\frac{a-2}{a-1}$

Eq$\left(1\right)$ line should pass through the point $(\frac{2-a^2}{a-1},\frac{a-2}{a-1})$

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

$2a-a^3+a-2=0$

$3a-a^3-2=0$

$a^3-3a+2=0$

Sum of roots $=0$

$\therefore$ , Sum of values of $a=0$