Question

If $\left|\begin{array}{ccc}a& b-c& c+b\\ a+c& b& c-a\\ a-b& a+b& c\end{array}\right|=0$, then the line $ax+by+c=0$ passes through the fixed point which is

• A. $(1,2)$
• B. $(1,1)$
• C. $(-2,1)$
• D. $(1,0)$

Answer 1

The correct option is B $(1,1)$

$\left|\begin{array}{ccc}a& b-c& c+b\\ a+c& b& c-a\\ a-b& b+a& c\end{array}\right|=0$
$C_1\to a{C}_1$
$\frac{1}{a}\left|\begin{array}{ccc}{a}^2& b-c& c+b\\ a^2+ac& b& c-a\\ a^2-ab& b+a& c\end{array}\right|=0$
$C_1\to C_1+b{C}_2+c{C}_3$
$\frac{1}{a}\left|\begin{array}{ccc}{a}^2+b^2+c^2& b-c& c+b\\ a^2+b^2+c^2& b& c-a\\ a^2+b^2+c^2& b+a& c\end{array}\right|=0$
$\frac{a^2+b^2+c^2}{a}\left|\begin{array}{ccc}1& b-c& c+b\\ 1& b& c-a\\ 1& b+a& c\end{array}\right|=0$
$R_2\to R_2-R_1,R_3\to R_3-R_1$
$\frac{a^2+b^2+c^2}{a}\left|\begin{array}{ccc}1& b-c& c+b\\ 0& c& -(a+b)\\ 0& a+c& -b\end{array}\right|=0$
$\Rightarrow \frac{a^2+b^2+c^2}{a}[-bc+a^2+ac+bc+ab]=0$
$\Rightarrow (a+b+c)(a^2+b^2+c^2)=0$
$\Rightarrow a+b+c=0$
Hence, the line $ax+by+c=0$ passes through the point $(1,1)$