A line $L$ is drawn from the point $P\equiv(4,3)$ to meet the parallel lines $L_1: 3x + 4y + 15 =0$ and $L_2 = 3x + 4y + 5 =0$ at points $A$ and $B$ respectively.
From $A$, a line perpendicular to $L$ is drawn meeting the line $L_2$ at $C$ . Similarly from point $B$, a line perpendicular to $L$ is drawn meeting the line $L_1$ at $D$. A parallelogram $ACBD$ is formed.
Find the equations(s) of the line $L$ such that the area of the parallelogram $ACBD$ is least.
The distance $BE$ between the parallel lines $L_1$ and $L_2$ is easily found and is $2$.
Let the angle which the line $L$ makes with $L_1$ be $\theta$. Simple trigonometry tells you that $AE=2\cot\theta$ and $CE=2\tan\theta$. Then the area of $\Delta ABC$ becomes: $$area=\frac{1}{2} base \cdot height$$ $$\Delta=\frac{1}{2} BE\cdot AC=\frac{1}{2} BE\cdot (AE+CE)=2|\cot\theta+\tan\theta|$$ The area of paralleogram is twice of $\Delta$, so it will be minimum when $\Delta$ is minimum. It is evident that $\Delta$ is minimum at $\theta=\dfrac{\pi}{4}$. At this value of $\theta$, clearly $\Delta_{min}=4.$
Now use the formula for angle between two lines to calculate the slope of line $L$: $$\pm\tan\theta={m_1-m_2 \over 1+m_1m_2}=\pm1$$ Take $m_1$ as the slope of $L_1$ and solve the equation to get two values of $m_2$. To obtain both the equations of the line $L$, use point-slope form of line since it passes through $P\equiv(4,3)$.
Wait for the GIF to load, it looks cool. You can see how the area approaches the minimum value of 8 twice and goes to infinity when the line $L$ becomes perpendicular to $L_1$.