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