Mathematics - Straight Lines Question with Solution | TestHub

Equation of the line, which is perpendicular to the line, $3x+y=\lambda (\lambda \ne 0)$ and passing through origin, is given by $$\frac{x-0}{3}=\frac{y-0}{1}=r$$ For foot of perpendicular $$r=\frac{-((3\times 0)+(1\times 0)-\lambda )}{3^2+1^2}=\frac{\lambda }{10}$$ So, foot of perpendicular $P=\left(\frac{3\lambda }{10},\frac{\lambda }{10}\right)$ Given the line meets $\mathrm{X}$-axis at $A=\left(\frac{\lambda }{3},0\right)$ and meets $Y$-axis at $B=(0,\lambda )$ So, $$\begin{array}{rl}& BP=\sqrt{{\left(\frac{3\lambda }{10}\right)}^2+{\left(\frac{\lambda }{10}-\lambda \right)}^2}\Rightarrow BP=\sqrt{\frac{9{\lambda }^2}{100}+\frac{81{\lambda }^2}{100}}\\ & \Rightarrow BP=\sqrt{\frac{90{\lambda }^2}{100}}\\ & \text{ Now, }PA=\sqrt{{\left(\frac{\lambda }{3}-\frac{3\lambda }{10}\right)}^2+{\left(0-\frac{\lambda }{10}\right)}^2}\\ & \Rightarrow PA\sqrt{\frac{{\lambda }^2}{900}+\frac{{\lambda }^2}{100}}\Rightarrow PA=\sqrt{\frac{10{\lambda }^2}{900}}\end{array}$$ Therefore $BP:PA=3:1$