Question

The foot of the perpendicular drawn from the origin, on the line, $3x+y=\lambda (\lambda \ne 0)$ is $P$. If the line meets x-axis at $A$ and y-axis at $B$, then the ratio $BP:PA$ is

• A. $9:1$
• B. $1:3$
• C. $1:9$
• D. $3:1$

Answer 1

The correct option is A $9:1$

Let (x,y) be foot of perpendicular drawn to the point $(x_1,y_1)$ on the line $ax+by+c=0$

Relation :$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-(a{x}_1+b{y}_1+c{z}_1)}{a^2+b^2}$

Here $(x_1,y_1)$=(0,0)

given line is: $3x+y-\lambda =0$

$\frac{x-0}{3}=\frac{y-0}{1}=\frac{-((3\times 0)+(1\times 0)-\lambda )}{3^2+1^2}$

$x=\frac{3\lambda }{10}$ and $y=\frac{\lambda }{10}$

Hence foot of perpendicular $P=(\frac{3\lambda }{10},\frac{\lambda }{10})$

Line meets X-axis at $A=(\frac{\lambda }{3},0)$

and meets Y-axis at $B=(0,\lambda )$

$BP=\sqrt{{\left(\frac{3\lambda }{10}\right)}^2+{(\frac{\lambda }{10}-\lambda )}^2}$

$\Rightarrow BP=\sqrt{\frac{9{\lambda }^2}{100}+\frac{81{\lambda }^2}{100}}$

$\therefore BP=\sqrt{\frac{90{\lambda }^2}{100}}$

$AP=\sqrt{{(\frac{\lambda }{3}-\frac{3\lambda }{10})}^2+{(0-\frac{\lambda }{10})}^2}$

$\Rightarrow AP=\sqrt{\frac{{\lambda }^2}{900}+\frac{{\lambda }^2}{100}}$

$\therefore AP=\sqrt{\frac{10{\lambda }^2}{900}}$

$\therefore BP:AP=9:1$

Hence,correct option is 'A'.