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(A) The equation of the line $3x + y = \lambda$ can be written in intercept form as $\frac{x}{\lambda/3} + \frac{y}{\lambda} = 1$. Thus,the coordinate

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$9 : 1$

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(A) The equation of the line $3x + y = \lambda$ can be written in intercept form as $\frac{x}{\lambda/3} + \frac{y}{\lambda} = 1$. Thus,the coordinates of $A$ are $(\frac{\lambda}{3}, 0)$ and $B$ are $(0, \lambda)$. The foot of the perpendicular $P$ from the origin $(0,0)$ to the line $ax + by + c = 0$ is given by $(\frac{-ac}{a^2+b^2}, \frac{-bc}{a^2+b^2})$. Here,$3x + y - \lambda = 0$,so $a=3, b=1, c=-\lambda$. $P = (\frac{-3(-\lambda)}{3^2+1^2}, \frac{-1(-\lambda)}{3^2+1^2}) = (\frac{3\lambda}{10}, \frac{\lambda}{10})$. Now,we calculate the distances $BP$ and $PA$: $BP^2 = (\frac{3\lambda}{10} - 0)^2 + (\frac{\lambda}{10} - \lambda)^2 = \frac{9\lambda^2}{100} + \frac{81\lambda^2}{100} = \frac{90\lambda^2}{100} = \frac{9\lambda^2}{10}$. $PA^2 = (\frac{\lambda}{3} - \frac{3\lambda}{10})^2 + (0 - \frac{\lambda}{10})^2 = (\frac{10\lambda - 9\lambda}{30})^2 + \frac{\lambda^2}{100} = \frac{\lambda^2}{900} + \frac{9\lambda^2}{900} = \frac{10\lambda^2}{900} = \frac{\lambda^2}{90}$. Therefore,$\frac{BP^2}{PA^2} = \frac{9\lambda^2/10}{\lambda^2/90} = \frac{9}{10} 	imes 90 = 81$. Thus,$\frac{BP}{PA} = \sqrt{81} = 9$. So,the ratio $BP : PA$ is $9 : 1$.

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

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