The line x + y = p meets the x and y axes at | vedclass

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(D) The line $x + y = p$ intersects the axes at $A(p, 0)$ and $B(0, p)$. The area of $	riangle OAB = \frac{1}{2} 	imes p 	imes p = \frac{p^2}{2}$.L

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$3$

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(D) The line $x + y = p$ intersects the axes at $A(p, 0)$ and $B(0, p)$. The area of $	riangle OAB = \frac{1}{2} 	imes p 	imes p = \frac{p^2}{2}$.Let $Q$ divide $AB$ in the ratio $\lambda : 1$. The coordinates of $Q$ are $\left(\frac{p}{1+\lambda}, \frac{p\lambda}{1+\lambda}ight)$.Since $PQ \perp AB$,the slope of $PQ$ is $1$ (as the slope of $AB$ is $-1$).The equation of line $PQ$ is $y - \frac{p\lambda}{1+\lambda} = 1(x - \frac{p}{1+\lambda})$.Setting $x = 0$ to find $P$ on the $y$-axis,we get $y = \frac{p\lambda - p}{1+\lambda} = \frac{p(\lambda - 1)}{\lambda + 1}$.Area of $	riangle APQ = \frac{1}{2} 	imes AQ 	imes PQ$. Using the coordinates,the area is $\frac{p^2 \lambda}{2(\lambda + 1)^2}$.Given $\frac{	ext{Area}(	riangle APQ)}{	ext{Area}(	riangle OAB)} = \frac{3}{8}$ $\Rightarrow \frac{p^2 \lambda / 2(\lambda + 1)^2}{p^2 / 2} = \frac{3}{8}$ $\Rightarrow \frac{\lambda}{(\lambda + 1)^2} = \frac{3}{8}$.$8\lambda = 3(\lambda^2 + 2\lambda + 1) \Rightarrow 3\lambda^2 - 2\lambda + 3 = 0$. This quadratic has no real roots. Re-evaluating the geometry,if $Q$ divides $AB$ in ratio $k:1$,then $\frac{AQ}{BQ} = k$. The ratio of areas is $\frac{k}{(k+1)^2} = \frac{3}{8}$ is incorrect. The correct ratio is $\frac{AQ}{BQ} = 3$ or $1/3$.

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