Question

Let $a,b,c$ be real numbers with $a^2+b^2+c^2=1$ Show that 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 straight line.

Answer 1

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