Question

Let a, b, c be real numbers with $a^2+b^2+c^2=1$ and $a\ne 0$. Then the equation $\left|\begin{array}{ccc}ax-by-c& bx+ay& cx+a\\ bx+ay& -ax+by-c& cy+b\\ cx+a& cy+b& -ax-by+c\end{array}\right|=0$ represents:

• A. A circle
• B. A pair of straight lines
• C. A straight line
• D. A parabola

Answer 1

The correct option is C A straight line

Given,
$\left|\begin{array}{ccc}ax-by-c& bx+ay& cx+a\\ bx+ay& -ax+by-c& cy+b\\ cx+a& cy+b& -ax-by+c\end{array}\right|=0$
Applying $C_1\to a{C}_1+b{C}_2+c{C}_3$

$\Rightarrow \left|\begin{array}{ccc}(a^2+b^2+c^2)x& bx+ay& cx+a\\ (a^2+b^2+c^2)y& -ax+by-c& cy+b\\ (a^2+b^2+c^2)& cy+b& -ax-by+c\end{array}\right|=0$
$\because a^2+b^2+c^2=1$ we have
$\left|\begin{array}{ccc}x& bx+ay& cx+a\\ y& -ax+by-c& cy+b\\ 1& cy+b& -ax-by+c\end{array}\right|=0$
Applying $C_2\to C_2-b{C}_1,C_3\to C_3-c{C}_1$
$\Rightarrow \left|\begin{array}{ccc}x& ay& a\\ y& -ax-c& b\\ 1& cy& -ax-by\end{array}\right|=0$
Applying $R_3\to R_3+x{R}_1+y{R}_2$, we get
$\Rightarrow \left|\begin{array}{ccc}x& ay& a\\ y& -ax-c& b\\ x^2+y^2+1& 0& 0\end{array}\right|=0$

$\Rightarrow (x^2+y^2+1)(aby+a^{2}x+ac)=0\Rightarrow ax+by+c=0$
Clearly, it represents a straight line.