The foot of the perpendicular drawn from the origin, on the line, 3x+y= (0) is P.

Question

The foot of the perpendicular drawn from the origin, on the line, $3x+y=\lambda\ (\lambda
e0)$ is $P$. If the line meets $x$-axis at $A$ and $y$-axis at $B$, then the ratio $BP:PA$ is:

Accepted Answer

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}}} $

Solutions · 1