Find equation of line(s) bisecting perimeter and area of triangle formed by $xy$-axes and $6x+8y=48$

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.)

There are three conditions possible :

1. $L$ intersects $OA$ and $AB$.

Area of $displaystyletriangle ACD=frac{1}{2}(pq)sin37^circ=12implies pq=40$. Also, $p+q=12$ for perimeter bisection.

Solve for $p,q$.

Do the same for other two cases

2. $L$ intersects $OB$ and $AB$.

3. $L$ intersects $OA$ and $OB$.

As corrected by @Moko19, also check the other three cases, too which I didn't mentioned in the diagram.