First, we find the coordinates of the points $P$ and $Q$ in terms of $k$ by finding the points of intersection of the graphs with equations $y=x^{2}$ and $y=3kx+4k^{2}$. Equating values of $y$, we obtain $x^{2}=3kx+4k^{2}$ or $x^{2}-3kx-4k^{2}=0$. We rewrite the left side as $x^{2}-4kx+kx+(-4k)(k)=0$ which allows us to factor and obtain $(x-4k)(x+k)=0$ and so $x=4k$ or $x=-k$. Since $k>0$, $P$ is in the second quadrant and $Q$ is in the first quadrant, then $P$ has $x$-coordinate $-k$ (which is negative). Since $P$ lies on $y=x^{2}$, then its $y$-coordinate is $(-k)^{2}=k^{2}$ and so the coordinates of $P$ are $(-k, k^{2})$. Since $Q$ lies on $y=x^{2}$ and has $x$-coordinate $4k$, then its $y$-coordinate is $(4k)^{2}=16k^{2}$ and so the coordinates of $Q$ are $(4k, 16k^{2})$. Our next step is to determine the area of $riangle OPQ$ in terms of $k$. Since the area of $riangle OPQ$ is numerically equal to 80, this will give us an equation for $k$ which will allow us to find the slope of the line. To find the area of $riangle OPQ$ in terms of $k$, we drop perpendiculars from $P$ and $Q$ to $S$ and $T$, respectively, on the $x$-axis. The area of $riangle OPQ$ is equal to the area of trapezoid $PSTQ$ minus the areas of $riangle PSO$ and $riangle QTO$. Trapezoid $PSTQ$ has parallel bases $SP$ and $TQ$ and perpendicular height $ST$. Since the coordinates of $P$ are $(-k, k^{2})$, then $SP=k^{2}$. Since the coordinates of $Q$ are $(4k, 16k^{2})$, then $TQ=16k^{2}$. Also, $ST=4k-(-k)=5k$. Thus, the area of trapezoid $PSTQ$ is $\frac{1}{2}(SP+TQ)(ST)=\frac{1}{2}(k^{2}+16k^{2})(5k)=\frac{85}{2}k^{3}$. $riangle PSO$ is right-angled at $S$ and so has area $\frac{1}{2}(SP)(SO)=\frac{1}{2}(k^{2})(0-(-k))=\frac{1}{2}k^{3}$. $riangle QTO$ is right-angled at $T$ and so has area $\frac{1}{2}(TQ)(TO)=\frac{1}{2}(16k^{2})(4k-0)=32k^{3}$. Combining these, the area of $riangle POQ$ equals $\frac{85}{2}k^{3}-\frac{1}{2}k^{3}-32k^{3}=10k^{3}$. Since this area equals 80, then $10k^{3}=80$ or $k^{3}=8$ and so $k=2$. This means that the slope of the line is $3k$ which equals 6.