A line $L$ through $(4,3)$ cuts two lines $L_1 \equiv 3x+4y+5=0$ and $L_2 \equiv 3x+4y+15=0$ at $A$ and $B$ respectively. From Point $A$ a perpendicular to $L$ is drawn to meet $L_2$ at $C$ and at $B$ a perpendicular to $L$ is drawn to meet $L_1$ at $D$. Now Paralellogram $ACBD$ is completed. Find Equation of $L$ such that area of Paralellogram is Least
My Try: we have height of Paralellogram as distance between lines $3x+4y+5=0$ and $3x+4y+15=0$ which is $2$
Since height is constant for area to be least base should be minimum.
so if we assume slope of $L$ as $m$ then its equation is $y-3=m(x-4)$ that is $mx-y+3-4m=0$
Solving $L$ and $L_1$ we get point $A$ and solving $L$ and $L_2$ we get point $B$ both in terms of $m$.
Measure distance $AB$ which is another height of paralellogram.
So area is $$\Delta =\frac{h_1h_2}{\sin \theta}$$
so $\Delta$ is a function of $m$ and use differentiation to minimize. But i ts very lengthy to solve in this approach. Any better way?