Explanation:
(B) : Family of straight lines given as-
\(x+(5 \lambda+1) y+1-3 \lambda=0 \ldots \text {.(i) }\)
\((5 \mu+2) x-3 y+3+6 \mu=0 \tag{ii}\)
From equation (i)
\((x+y+1)+\lambda(5 y-3)=0\)
\(\begin{array}{cl}\text { Which gives } x+y+1=0 \\ \text { and } 5 y-3=0\end{array}\)
Solving equations (iii) and (iv), we get-
Point \(\mathrm{P}\left(\frac{-8}{5}, \frac{3}{5}\right)\)
Similarly from equation (ii)
Which gives
\((2 x-3 y+3)+\mu(5 x+6)=0\)
\(2 x-3 y+3=0 \tag{v}\)
\(x=\frac{-6}{5} \tag{vi}\)
Solving equation (v) and (vi), we get the point \(\mathrm{Q}\left(\frac{-6}{5}, \frac{1}{5}\right)\)
Hence, point of concurrency are \(\mathrm{P}\left(\frac{-8}{5}, \frac{3}{5}\right)\) and
\(\mathrm{Q}\left(\frac{-6}{5}, \frac{1}{5}\right)\)
\(\therefore \quad \mathrm{PQ} =\sqrt{\left(\frac{-8}{5}+\frac{6}{5}\right)^{2}+\left(\frac{3}{5}-\frac{1}{5}\right)^{2}}\)
\(=\frac{\sqrt{8}}{5}=\frac{2 \sqrt{2}}{5}\)
\(\Rightarrow \mathrm{p} \cos \alpha-\mathrm{q} \sin \alpha=\sqrt{3}\) and \(\mathrm{p} \sin \alpha+\mathrm{q} \cos \alpha=-1\)
Adding and squaring, we get
\(\mathrm{p}^{2}+\mathrm{q}^{2}=3+1 \Rightarrow \mathrm{p}^{2}+\mathrm{q}^{2}=4\)
