QJEEM18Mae642: JEE Main 2018 Mathematics - Class 11 JEEM18Mae642 Solution: The foot of the perpendicular drawn from the origin, on the…

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

3x + y = $\lambda$ ( $\lambda$ $\ne$ 0) and passing through origin ,

is given by ${{x - 0} \over 3} = {{y - 0} \over 1} = r$

For foot of perpendicular

r = ${{ - \left( {\left( {3 \times 0} \right) + \left( {1 \times 0} \right) - \lambda } \right)} \over {{3^2} + {1^2}}}$ = ${\lambda \over {10}}$

So, foot of perpendicular P = $\left( {{{3\lambda } \over {10}},{\lambda \over {10}}} \right)$

Given the line meets X-axis where y = 0, so 3x + 0 = $\lambda$

$\Rightarrow$ x = ${\lambda \over 3}$

Hence, coordinates of A = $\left( {{\lambda \over 3},0} \right)$ and meets

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

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

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

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

Now, PA = $\sqrt {{{\left( {{\lambda \over 3} - {{3\lambda } \over {10}}} \right)}^2} + {{\left( {0 - {\lambda \over {10}}} \right)}^2}}$

$\Rightarrow$ $\,\,\,$ PA = $\sqrt {{{{\lambda ^2}} \over {900}} + {{{\lambda ^2}} \over {100}}} \Rightarrow PA$ = $\sqrt {{{10{\lambda ^2}} \over {900}}}$

Therefore BP : PA = 9 : 1