Question

If the straight lines $ax+by+c=0$ and $x\cos \alpha +y\sin \alpha +c=0$ encloses an angle $\frac{\pi }{4}$ between them and meet the striaght lie $x\sin \alpha -y\cos \alpha =0$ in the same point, then

• A. $a^2+b^2=c^2$
• B. $a^2+b^2=2$
• C. $a^2+b^2=2c^2$
• D. $a^2+b^2=4$

Answer 1

Answer: (B)

$a^2+b^2=2$

$L_1\to ar+by+c=0$

$L_2\to x\cos \alpha +y\sin \alpha ac=0$

we have

$a\cos \alpha +b\sin \alpha =\sqrt{a^2+b^2}\cos \left(\frac{n}{4}\right)$ or

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

but from

$x\cos \alpha +y\sin \alpha +c=0$

$x\sin \alpha -y\cos \alpha =0$

intersection point give us

$x=c\cos \alpha$

$y=c\sin \alpha$

and after substitution into $aax+by+c=0=1$

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

then

$a^2+b^2=2$

Ans $\left(B\right)$