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 $(1,1)$

Applying $C_1\to a{C}_1$ and then $C_1\to C_1+b{C}_2+c{C}_3,$ and taking $a^2+b^2+c^2$ common from $C_1,$ we get
$△=\frac{(a^2+b^2+c^2)}{a}\left|\begin{array}{ccc}1& b-c& c+b\\ 1& b& c-a\\ 1& b+a& c\end{array}\right|$

$R_2\to R_2-R_1,R_3\to R_3-R_1$

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

Expanding along $C_1$
$=\frac{(a^2+b^2+c^2)}{a}(-bc+a^2+ab+ac+bc)$

$=(a^2+b^2+c^2)(a+b+c)$

Hence, $△=0\Rightarrow a+b+c=0$
Therefore, line $ax+by+c=0$ passes through the fixed point $(1,1)$.