The equation of the line through $(3, 4)$ which cuts from the first quadrant a triangle of minimum area is

$\begin{array}{r}\text{ Slope }=\frac{-y}{x}=\frac{4}{3-x}\\ \text{ area }=\frac{1}{2}xy\\ =\frac{1}{2}x\left(\frac{4x}{x-3}\right)\\ \frac{d\left(\text{ area }\right)}{dx}=2\left(\frac{2x}{(x-3)}-\frac{x^2}{(x-3{)}^2}\right.\\ 0=2\left(\frac{x}{x-3}\right)\left(\frac{2(x-3)-x}{(x-3)}\right)\\ 0=\frac{2x(x-6)}{(x-3{)}^2}\\ x=0,6\end{array}$

$\begin{array}{r}\frac{d\left(\text{ area }\right)}{dx}<0\text{ If }0<x<6\\ \frac{d\left(\text{ area }\right)}{dx}>0\text{ If }x<0\text{ or }x>6\end{array}$

$\text{ So area is min when }x=6,y=8$

(3,4) is the mid point of the line segment.