Question

Let a, b, c be real numbers with $a^2+b^2+c^2=1$. 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. parabola
• B. straight line
• C. ellipse
• D. hyperbola

Answer 1

The correct option is B straight line

Given, $\left|\begin{array}{ccc}ax-by-c& bx+ay& cx+a\\ bx+ay& -ax+by-c& cy+b\\ cx+a& cy+b& -ax-by+c\end{array}\right|=0$
$\Rightarrow \frac{1}{a}\left|\begin{array}{ccc}{a}^{2}x-aby-ac& bx+ay& cx+a\\ abx+a^{2}y& -ax+by-c& cy+b\\ acx+a^2& cy+b& -ax-by+c\end{array}\right|=0$
Applying $C_1\to C_1+b{C}_2+c{C}_3$
$\Rightarrow \frac{1}{a}\left|\begin{array}{ccc}(a^2+b^2+c^2)x& ay+bx& cx+a\\ (a^2+b^2+c^2)y& by-c-ax& b+cy\\ a^2+b^2+c^2& b+cy& c-ax-by\end{array}\right|=0$
$\Rightarrow \frac{1}{a}\left|\begin{array}{ccc}x& ay+bx& cx+a\\ y& by-c-ax& b+cy\\ 1& b+cy& c-ax-by\end{array}\right|=0$
$[\because a^2+b^2+c^2=1]$
Applying $C_2\to C_2-b{C}_{1}and{C}_3\to C_3-c{C}_1$
$\Rightarrow \frac{1}{a}\left|\begin{array}{ccc}x& ay& a\\ y& -c-ax& b\\ 1& cy& -ax-by\end{array}\right|=0\Rightarrow \frac{1}{ax}\left|\begin{array}{ccc}{x}^2& axy& ax\\ y& -c-ax& b\\ 1& cy& -ax-by\end{array}\right|=0$
Applying $R_1\to R_1+y{R}_2+R_3$
$\Rightarrow \frac{1}{ax}\left|\begin{array}{ccc}{x}^2+y^2+1& 0& 0\\ y& -c-ax& b\\ 1& cy& -ax-by\end{array}\right|=0\Rightarrow \frac{1}{ax}\left[\right(x^2+y^2+1\left)\right\{(-c-ax)(-ax-by)-b\left(cy\right)\left\}\right]=0\Rightarrow \frac{1}{ax}\left[\right(x^2+y^2+1\left)\right(acx+bcy+a^2{x}^2+abxy-bcy\left)\right]=0\Rightarrow \frac{1}{ax}\left[\right(x^2+y^2+1\left)\right(acx+a^2{x}^2+abxy\left)\right]=0\Rightarrow \frac{1}{ax}\left[ax\right(x^2+y^2+1\left)\right(c+ax+by\left)\right]=0\Rightarrow (x^2+y^2+1)(ax+by+c)=0\Rightarrow ax+by+c=0$
which represents a straight line.