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

Straight Lines and Pair of Straight Lines PYQ — JEE Main 2018 · 15 Apr · Shift 2 · Q35 with Solution

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Correct answer: D

The line is $3x+y=\lambda, \qquad \lambda\ne 0.$

It meets the axes at:

On the $x$ -axis, $y=0$ : $3x=\lambda \Rightarrow x=\frac{\lambda}{3}.$ So, $A\left(\frac{\lambda}{3},0\right).$
On the $y$ -axis, $x=0$ : $y=\lambda.$ So, $B(0,\lambda).$

For the line $ax+by+c=0,$ the foot of perpendicular from $(0,0)$ is $\left(\frac{-ac}{a^2+b^2},\frac{-bc}{a^2+b^2}\right).$

Here, $3x+y-\lambda=0,$ so $a=3,\quad b=1,\quad c=-\lambda.$

Thus, $P=\left(\frac{-3(-\lambda)}{3^2+1^2},\frac{-1(-\lambda)}{3^2+1^2}\right)=\left(\frac{3\lambda}{10},\frac{\lambda}{10}\right).$

Now, $A\left(\frac{\lambda}{3},0\right), \qquad B(0,\lambda).$

Suppose $P$ divides $AB$ in the ratio $BP:PA=m:n.$

Then by section formula,

=\left(\frac{m\lambda}{3(m+n)},\frac{n\lambda}{m+n}\right).$$ Comparing with $$P=\left(\frac{3\lambda}{10},\frac{\lambda}{10}\right),$$ we get From the $y$-coordinate: $$\frac{n\lambda}{m+n}=\frac{\lambda}{10} \Rightarrow \frac{n}{m+n}=\frac{1}{10} \Rightarrow 10n=m+n \Rightarrow m=9n.$$ Hence, $$BP:PA=m:n=9:1.$$ --- 4. **Check options** The correct option is $$\boxed{9:1}.$$ So, **Option D** is correct.