JEE Main 2018 — Mathematics Coordinate Geometry

Equation of the line, which is perpendicular to the line, $3 x+y=\lambda(\lambda \neq 0)$3x+y=λ(λ=0) and passing through origin, is given by $$\frac{x-0}{3}=\frac{y-0}{1}=r$$3x−0​=1y−0​=r For foot of perpendicular $$r=\frac{-((3 \times 0)+(1 \times 0)-\lambda)}{3^2+1^2}=\frac{\lambda}{10}$$r=32+12−((3×0)+(1×0)−λ)​=10λ​ So, foot of perpendicular $P=\left(\frac{3 \lambda}{10}, \frac{\lambda}{10}\right)$P=(103λ​,10λ​) Given the line meets $\mathrm{X}$X-axis at $A=\left(\frac{\lambda}{3}, 0\right)$A=(3λ​,0) and meets $Y$Y-axis at $B=(0, \lambda)$B=(0,λ) So, $$\begin{aligned} &B P=\sqrt{\left(\frac{3 \lambda}{10}\right)^2+\left(\frac{\lambda}{10}-\lambda\right)^2} \Rightarrow B P=\sqrt{\frac{9 \lambda^2}{100}+\frac{81 \lambda^2}{100}} \\ &\Rightarrow B P=\sqrt{\frac{90 \lambda^2}{100}} \\ &\text { Now, } P A=\sqrt{\left(\frac{\lambda}{3}-\frac{3 \lambda}{10}\right)^2+\left(0-\frac{\lambda}{10}\right)^2} \\ &\Rightarrow P A \sqrt{\frac{\lambda^2}{900}+\frac{\lambda^2}{100}} \Rightarrow P A=\sqrt{\frac{10 \lambda^2}{900}} \end{aligned}$$​BP=(103λ​)2+(10λ​−λ)2​⇒BP=1009λ2​+10081λ2​​⇒BP=10090λ2​​ Now, PA=(3λ​−103λ​)2+(0−10λ​)2​⇒PA900λ2​+100λ2​​⇒PA=90010λ2​​​ Therefore $B P: P A=3: 1$BP:PA=3:1