Mathematics - Straight Lines Question with Solution | TestHub

Let area of $△PBC=x.\Rightarrow \frac{x}{5+\lambda }=\frac{8}{22-\lambda }$ and $\frac{x}{30-\lambda }=\frac{5}{\lambda }\Rightarrow \frac{\lambda +5}{30-\lambda }=\frac{(22-\lambda )5}{8\lambda }\Rightarrow 8{\lambda }^2+40\lambda =5({\lambda }^2-$ $52\lambda +660)\Rightarrow {\lambda }^2+100\lambda -1100=0\Rightarrow (\lambda +110)(\lambda -10)=0\Rightarrow \lambda =10\Rightarrow x=\frac{(30-\lambda )5}{\lambda }=\frac{(30-10)\times 5}{10}=10$ square units. Let lambda be area APF. Put x=area BPC. Then area APC=30-lambda and area ABP=5+lambda. The area ratios use distances from A,B to CF, and from A,C to BE, respectively. Thus lambda/5=(30-lambda)/x and (22-lambda)/8=(5+lambda)/x. Positive areas exclude lambda=-110. Hence lambda=10 and x=10, option C.