Question

If the straight lines ax + by + p = 0 and $xcos\alpha +ysin\alpha -p=0$ include an angle $\frac{\pi }{4}$ between them and meet the
straight line $xsin\alpha -ycos\alpha =0$ in the same point, then the value of $a^2+b^2$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

It is given that the lines ax + by + p = 0 and $xcos\alpha +ysin\alpha =p$ are inclined at an angle $\frac{\pi }{4}$
Therefore $tan\frac{\pi }{4}=\frac{-\frac{a}{b}+\frac{cos\alpha }{sin\alpha }}{1+\frac{acos\alpha }{bsin\alpha }}\Rightarrow acos\alpha +bsin\alpha =-asin\alpha +bcos\alpha$ ....(i)
It is given that the lines ax + by + p = 0, $xcos\alpha +ysin\alpha -p=0$ and $xsin\alpha -ycos\alpha =0$ are concurrent.
$\therefore \left|\begin{array}{ccc}a& b& p\\ cos\alpha & sin\alpha & -p\\ sin\alpha & -cos\alpha & 0\end{array}\right|\Rightarrow -apcos\alpha -bpsin\alpha -p=0\Rightarrow -acos\alpha -bsin\alpha =1\Rightarrow acos\alpha +bsin\alpha =-1From\left(i\right)and\left(ii\right),-asin\alpha +bcos\alpha =-1\dots \dots \left(ii\right)From\left(ii\right)and\left(iii\right)(acos\alpha +bsin\alpha {)}^2+(-acos\alpha +bsin\alpha {)}^2=2\Rightarrow a^2+b^2=2.$