TITLE: Find equation of line(s) bisecting perimeter and area of triangle formed by $xy$-axes and $6x+8y=48$
QUESTION [5 upvotes]: The line $6x+8y=48$ intersects the $x$-axis in point $A$ and the $y$-axis in point $B$. A line $L$ bisects the area and the perimeter of the triangle $OAB$,where $O$ is origin.
Find possible equation(s) of $L$.
My Attempt
I feel there can be three lines but how to proceed.

REPLY [3 votes]: Let the line $L$ be $y= k x +b$ and it intersects with $6x+8y=48$ at $(p,q)$, where
$$p= \frac{24-4b}{3+4k },\>\>\>\>\> q= \frac{24k+3b}{3+4k } $$
Given the equal area and perimeter, establish the equations below
$$(6-b)p=bp+8q$$
$$(6-b)+\sqrt{p^2+(6-q)^2} = b+8+\sqrt{(8-p)^2+q^2}$$
Solve the system of equations above to obtain $b=\sqrt6$ and $k=1-\sqrt{\frac32}$. Thus, the bisecting line $L$ is
$$y= \left(1-\sqrt{\frac32}\right)x+\sqrt6$$
(Note that above equations corresponds to $L$ intersecting the vertical leg and the hypotenuse; other configurations do not yield valid solutions.)