Question

If the equation $6x^2-11xy-10y^2-2x-14y+c=0$ represents a pair of lines, then which of the following shows the correct lines and the angle between them?

• A. $2x+5y-2=0$ & $3x-2y+3=0;\pm {\tan }^{-1}\left(\frac{17}{4}\right)$
• B. $2x-5y+2=0$ & $3x+2y+3=0;\pm {\tan }^{-1}\left(\frac{19}{4}\right)$
• C. $2x-5y-2=0$ & $3x+2y+2=0;\pm {\tan }^{-1}\left(\frac{9}{2}\right)$
• D. $2x+5y+2=0$ & $3x-2y+3=0;\pm {\tan }^{-1}\left(\frac{17}{4}\right)$

Answer 1

The correct option is B $2x-5y+2=0$ & $3x+2y+3=0;\pm {\tan }^{-1}\left(\frac{19}{4}\right)$

Given equation represents a pair of lines when
$\left|\begin{array}{ccc}6& -1& -\frac{11}{2}\\ -1& -10& -7\\ -\frac{11}{2}& -7& c\end{array}\right|=0$
$\Rightarrow 6(-10c-49)+1(-c-\frac{77}{2})-\frac{11}{2}(7-55)=0$
$\Rightarrow c=-4$
Hence equation of pair of lines is,
$6x^2-11xy-10y^2-2x-14y-4=0$
$\Rightarrow (2x-5y-2)(3x+2y+3)=0$
$\Rightarrow 2x-5y-2=0$ and $3x+2y+2=0$
And angle between them is,
$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}=\frac{2\sqrt{\frac{121}{4}+60}}{6-10}=\pm \frac{19}{4}=$