Question

If the straight line $ax+by+p=0$ and $xcos\alpha +ysin\alpha =p$ enclosed an angle of $\frac{\pi }{4}$ and the line $xsin\alpha -ycos\alpha =0$ meets them at the same point , the $a^2+b^2$ is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (C)

$2$

Given lines: $ax+by+p=0$ $m_1=\frac{-a}{b}$

$xcos\alpha +ysin\alpha =p$ $m_2=\frac{-cos\alpha }{sin\alpha }$

Both lines enclose an angle of ${45}^{\circ }$, Thus,

$tan{45}^{\circ }=\frac{|m_1-m_2|}{|1+m_1{m}_2|}$

$\Rightarrow |1+m_1{m}_2|=|m_1-m_2|$

$\Rightarrow |bsin\alpha +acos\alpha |=|asin\alpha -bcos\alpha |$

Now, squaring both sides we get:

$b^{2}si{n}^{2}A+a^{2}co{s}^{2}A+2absinAcosA=a^{2}si{n}^{2}A+b^{2}co{s}^{2}A-2absinAcosA$ $\to ⟨1⟩$

Also, given that 3 lines are $concurrent$(meets at same point).

Point of intersection of line: $xcos\alpha +ysin\alpha =p$ & $xsin\alpha -ycos\alpha =0$ is:

$\Rightarrow (pcos\alpha ,psin\alpha )$

As line $ax+by+p=0$ also passes through this point so,

$a\left(pcos\alpha \right)+b\left(psin\alpha \right)=-p$

$acos\alpha +bsin\alpha =-1$

Squaring both sides, we get:

$b^{2}si{n}^{2}A+a^{2}co{s}^{2}A+2absinAcosA=1$ $\to ⟨2⟩$

Now, Using $⟨2⟩in⟨1⟩$ we get:

$a^{2}si{n}^{2}A+b^{2}co{s}^{2}A-2absinAcosA=1$ $\to ⟨3⟩$

Adding $⟨2⟩$ & $⟨3⟩$:

$a^2+b^2=2$

Hence, Option $\left(C\right)$ is correct.