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

Correct answer: D. 3: 1 Equation of the line, which is perpendicular to the line, 3 x+y=λ(λ ≠ 0) and passing through origin, is given by (x-0)/(3)=(y-0)/(1)= r For foot of perpendicular r= (-((3 × 0)+(1 × 0)-λ))/(3^2+1^2)=(λ)/(10) So, foot of perpendicular P= ((3 λ)/(10), (λ)/(10)) Given the line meets X -axis at A= ((λ)/(3), 0) and meets Y -axis at B=(0, λ) So, aligned &B P= ((3 λ)/(10))^2+((λ)/(10)-λ)^2 P= (9 λ^2)/(100)+(81 λ^2)/(100) & B P= (90 λ^2)/(100) Now, P A= ((λ)/(3)-(3 λ)/(10))^2+(0-(λ)/(10))^2 & P A (λ^2)/(900)+(λ^2)/(100) P A= (10 λ^2)/(900) aligned Therefore B P: P A=3: 1

JEE Main 201815 AprilEvening ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Evening Shift
Mode: Online
Subject: Mathematics
Chapter: Straight Lines

JEE Main 2018 · 15 April · Evening Shift · Online · Mathematics · Straight Lines

The foot of the perpendicular drawn from the origin, on the line, 3...

The foot of the perpendicular drawn from the origin, on the line,

3 x+y=\lambda(\lambda \neq 0)

P

. If the line meets

x

-axis at

A

and

y

-axis at

B

, then the ratio

B P

: P A

Options

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

Correct answer

D. $3: 1$

Step-by-step solution

Equation of the line, which is perpendicular to the line,

3 x+y=\lambda(\lambda \neq 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{aligned}&B P=\sqrt{\left(\frac{3 \lambda}{10}\right)^2+\left(\frac{\lambda}{10}-\lambda\right)^2} \RightarrowB 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}

Therefore

B P: P A=3: 1

Practice Straight Lines on Quantrex Academy →

More from Straight Lines