Question

If the straight lines $ax+by+p=0$ and $x\cos a+y\sin a=p$ are inclined at an angle $\frac{\pi }{4}$ and concurrent with the straight line $x\sin a-y\cos a=0$ , then the value of $a^2+b^2$ is

• A. $0$
• B. $1$
• C. $2$
• D. none of these

Answer 1

The correct option is C $2$

$\tan \frac{\pi }{4}=\left|\frac{-\frac{a}{b}-(-\frac{\cos \alpha }{\sin \alpha })}{1+(-\frac{a}{b})(-\frac{\cos \alpha }{\sin \alpha })}\right|$

$\Rightarrow |b\sin \alpha +a\cos \alpha |=|b\cos \alpha -a\sin \alpha |$

$\Rightarrow b^2{\sin }^2\alpha +a^2{\cos }^2\alpha +2ab\sin \alpha \cos \alpha =b^2{\cos }^2\alpha +a^2{\sin }^2\alpha -2ab\sin \alpha \cos \alpha$ .....$\left(1\right)$

Solving $x\cos \alpha +y\sin \alpha =c$ and $x\sin \alpha -y\cos \alpha =0$, we get

$x\sin \alpha =y\cos \alpha$

or $x=\frac{y\cos \alpha }{\sin \alpha }$

substituting $x=\frac{y\cos \alpha }{\sin \alpha }$ in $x\cos \alpha +y\sin \alpha =c$ we get

$x\cos \alpha +y\sin \alpha =c$

$\Rightarrow \frac{y\cos \alpha }{\sin \alpha }\cos \alpha +y\sin \alpha =c$

$\Rightarrow y{\cos }^2\alpha +y{\sin }^2\alpha =c\sin \alpha$

$\Rightarrow y({\cos }^2\alpha +{\sin }^2\alpha )=c\sin \alpha$

$\Rightarrow y=c\sin \alpha$

substitute $y=c\sin \alpha$ in $x\sin \alpha -y\cos \alpha =0$ we get

$x\sin \alpha =y\cos \alpha \Rightarrow x\sin \alpha =c\sin \alpha \cos \alpha$

$\therefore x=c\cos \alpha$

The three lines are concurrent.

$\therefore ac\cos \alpha +bc\sin \alpha +c=0$

$\Rightarrow a\cos \alpha +b\sin \alpha =-1$

$\Rightarrow a^2{\cos }^2\alpha +b^2{\sin }^2\alpha +2ab\sin \alpha \cos \alpha =1$ ......$\left(2\right)$

Fro\alpha $\left(1\right)$ and $\left(2\right)$

$b^2{\cos }^2\alpha +a^2{\sin }^2\alpha -2ab\sin \alpha \cos \alpha =1$ ......$\left(3\right)$

Adding $\left(2\right)$ and $\left(3\right)$,

$a^2{\cos }^2\alpha +b^2{\sin }^2\alpha +2ab\sin \alpha \cos \alpha +b^2{\cos }^2\alpha +a^2{\sin }^2\alpha -2ab\sin \alpha \cos \alpha =1+1$

$a^2{\cos }^2\alpha +a^2{\sin }^2\alpha +b^2{\sin }^2\alpha +b^2{\cos }^2\alpha =2$

$\therefore a^2+b^2=2$