JEE Main 2013 — Mathematics Coordinate Geometry

For $\mathrm{L}_1$L1​, $$\begin{aligned} & x=\sqrt{\lambda} y+(\sqrt{\lambda}-1) \Rightarrow y=\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}} \\ & z=(\sqrt{\lambda}-1) y+\sqrt{\lambda} \Rightarrow y=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1} \end{aligned}$$​x=λ​y+(λ​−1)⇒y=λ​x−(λ​−1)​z=(λ​−1)y+λ​⇒y=λ​−1z−λ​​​ From (i) and (ii) $$\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}=\frac{y-0}{1}=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}$$λ​x−(λ​−1)​=1y−0​=λ​−1z−λ​​ The equation (A) is the equation of line $\mathrm{L}_1$L1​. Similarly equation of line $\mathrm{L}_2$L2​ is $$\frac{x-(1-\sqrt{\mu})}{\sqrt{\mu}}=\frac{y-0}{1}=\frac{z-\sqrt{\mu}}{1-\sqrt{\mu}}$$μ​x−(1−μ​)​=1y−0​=1−μ​z−μ​​ Since $\mathrm{L}_1 \perp \mathrm{L}_2$L1​⊥L2​, therefore $$\begin{aligned} & \sqrt{\lambda} \sqrt{\mu}+1 \times 1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0 \\ & \Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \sqrt{\lambda}=-\sqrt{\mu} \\ & \Rightarrow \lambda=\mu \end{aligned}$$​λ​μ​+1×1+(λ​−1)(1−μ​)=0⇒λ​+μ​=0⇒λ​=−μ​⇒λ=μ​