Question

If the equation $3x^2+kxy-10y^2+7x+19y=6$ represents a pair of lines, find the value of $k$

• A. $k=1$, or $\frac{127}{6}$
• B. $k=-1$, or $-\frac{127}{6}$
• C. $k=6$, or $\frac{133}{6}$
• D. $k=-6$, or $-\frac{133}{6}$

Answer 1

Answer: (B)

$k=-1$, or $-\frac{127}{6}$

This equation represent pair of lines when

$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$

$\left|\begin{array}{ccc}3& \frac{7}{2}& -\frac{k}{2}\\ \frac{7}{2}& -10& \frac{19}{2}\\ -\frac{k}{2}& \frac{19}{2}& -6\end{array}\right|=0$
$\Rightarrow 3(60-\frac{361}{4})-\frac{7}{2}(-21-\frac{19k}{4})-\frac{k}{2}(\frac{133}{4}-5k)=0$
$\Rightarrow k=-1,-\frac{127}{6}$