A) \[p=12,\,\,q=-1\]
B) \[p=-12,\,\,q=1\]
C) \[p=12,\,\,q=1\]
D) \[p=1,\,\,q=1\]
Correct Answer: C
Solution :
The second degree equation will represent a pair of perpendicular straight lines, if \[\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|=0\] and\[a+b=0\] Given pair of line is \[12{{x}^{2}}+7xy-p{{y}^{2}}-18x+qy+6=0\] \[\therefore \] \[\left| \begin{matrix} 12 & 7/2 & -9 \\ 7/2 & -p & q/2 \\ -9 & q/2 & 6 \\ \end{matrix} \right|=0\] and \[12-p=0\] \[\Rightarrow \,\,\,\,\,\,p=12\] \[\therefore \,\,\,\,\,\,\,\,\,\,\left| \begin{matrix} 12 & 7/2 & -9 \\ 7/2 & -12 & 1/2 \\ -9 & q/2 & 6 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[12\left( -72+\frac{{{q}^{2}}}{4} \right)-\frac{7}{2}\left( -21+\frac{9q}{2} \right)\] \[-9\left( \frac{7q}{4}-108 \right)=0\] \[\Rightarrow \] \[-864-3{{q}^{2}}-\frac{147}{2}-\frac{63q}{4}\] \[-\frac{63q}{4}+972=0\Rightarrow q=1\]
You need to login to perform this action.
You will be redirected in
3 sec
