, 2022

Line $L_1$ has equation $y=\frac32x+k$, and thus has slope $\frac32$. Since $L_2$ is perpendicular to $L_1$, then its slope is $-\frac23$. Since $L_1$ has equation $y=\frac32x+k$, its $y$-intercept is $k$. Line $L_2$ has the same $y$-intercept as $L_1$ (which is at $P(0,k)$). Thus, $L_2$ has slope $-\frac23$ and $y$-intercept $k$, and so it has equation $y=-\frac23x+k$. Since $L_2$ intersects the $x$-axis at $Q$, the $x$-coordinate of point $Q$ is the $x$-intercept of $L_2$. Setting $y=0$ in the equation for $L_2$ and solving for $x$, we get $0=-\frac23x+k$ or $\frac23x=k$, and so $x=\frac{3k}{2}$. Written in terms of $k$, the $x$-coordinate of point $Q$ is $\frac{3k}{2}$. From part (b), the coordinates of $P$ are $(0,k)$, and the coordinates of $Q$ are $\left(\frac{3k}{2},0\right)$. To determine an expression for the area of $\triangle PQR$, we first need to determine the coordinates of point $R$. $L_3$ is parallel to $L_1$ and thus has slope $\frac32$ and equation $y=\frac32x+b$, for some $y$-intercept $b$. Line $L_3$ passes through point $Q\left(\frac{3k}{2},0\right)$, and so $0=\frac32\left(\frac{3k}{2}\right)+b$ or $b=-\frac{9k}{4}$. Therefore the $y$-intercept of $L_3$ is $-\frac{9k}{4}$ and so $R$ has coordinates $\left(0,-\frac{9k}{4}\right)$. If we call the origin $O(0,0)$, then the area of $\triangle PQR$ is given by $\frac12\times PR\times OQ$, since height $OQ$ is perpendicular to the base $PR$. Since $PR=k-\left(-\frac{9k}{4}\right)=\frac{13k}{4}$ and $OQ=\frac{3k}{2}$, then the area of $\triangle PQR$ is $\frac12\times\frac{13k}{4}\times\frac{3k}{2}=\frac{39k^2}{16}$. The area of $\triangle PQR$ is 351, and so $\frac{39k^2}{16}=351$ or $k^2=\frac{351\times16}{39}$ or $k^2=144$, and so $k=12$ (since $k>0$).