Question

If $3 x^{2}-11 x y+10 y^{2}-7 x+13 y +k=0$ denotes a pair of straight lines, then the point of intersection of the lines is

• A. (1, 3)
• B. (3, 1)
• C. (-3, 1)
• D. (1, -3)

Answer 1

Answer: (B)

The pair of straight lines is
$3 x^{2}-11 x y+10 y^{2}-7 x+13 y+ k=0$
then on compairing with
$a x^{2}+2 h x y+ b y^{2}+2 g x+2 f y+ c=0$
$\Rightarrow
\begin{cases}
a=3, h=-\frac{11}{2}, b=10 \\
g=-\frac{7}{2}, f=\frac{13}{2}, c=k
\end{cases}$
Then, the point of intersection of the lines is
$\left[\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right]$
$=\left[\frac{\frac{-143}{4}+\frac{70}{2}}{30-\frac{121}{4}}, \frac{\frac{77}{4}-\frac{39}{2}}{30-\frac{121}{4}}\right]$
$=\left[\frac{-3}{120-121}, \frac{77-78}{120-121}\right]$
$=\left(\frac{-3}{-1}, \frac{-1}{-1}\right)=(3,1)$