The foot of the perpendicular drawn from the origin, on the line, $$3x + y = \lambda (\lambda \neq 0)$$ is $$P$$. If the line meets x-axis at $$A$$ and y-axis at $$B$$, then the ratio $$BP : PA$$ is
Let (x,y) be foot of perpendicular drawn to the point $$(x_1,y_1)$$ on the line $$ax+by+c=0$$
Relation :$$\dfrac{x-x_{1}}{a}=\dfrac{y-y_{1}}{b}=\dfrac{-\left ( ax_{1}+by_{1}+cz_{1} \right )}{a^{2}+b^{2}}$$
Here $$(x_1,y_1)$$=(0,0)
given line is: $$3x+y-\lambda=0$$
$$\dfrac{x-0}{3}=\dfrac{y-0}{1}=\dfrac{-\left ( \left ( 3\times 0 \right )+\left ( 1\times 0 \right )-\lambda \right )}{3^{2}+1^{2}}$$
$$x=\dfrac{3\lambda }{10}$$ and $$y=\dfrac{\lambda }{10}$$
Hence foot of perpendicular $$P=\left ( \dfrac{3\lambda }{10},\dfrac{\lambda }{10} \right )$$
Line meets X-axis at $$A=\left ( \dfrac{\lambda }{3},0 \right )$$
and meets Y-axis at $$B=\left ( 0,\lambda \right )$$
$$BP=\sqrt{\left ( \dfrac{3\lambda }{10} \right )^{2}+\left ( \dfrac{\lambda }{10}-\lambda \right )^{2}}$$
$$\Rightarrow BP=\sqrt{\dfrac{9\lambda ^{2}}{100}+\dfrac{81\lambda ^{2}}{100}}$$
$$\therefore BP=\sqrt{\dfrac{90\lambda ^{2}}{100}}$$
$$AP=\sqrt{\left ( \dfrac{\lambda }{3}-\dfrac{3\lambda }{10} \right )^{2}+\left ( 0-\dfrac{\lambda }{10} \right )^{2}}$$
$$\Rightarrow AP=\sqrt{\dfrac{\lambda ^{2}}{900}+\dfrac{\lambda ^{2}}{100}}$$
$$\therefore AP=\sqrt{\dfrac{10\lambda ^{2}}{900}}$$
$$\therefore BP:AP=9:1$$
Hence,correct option is 'A'.
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The foot of the perpendicular from the point $$(3, 4)$$ on the line $$3x-4y+5=0$$ is:
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