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 $\displaystyle\triangle ACD=\frac{1}{2}(pq)\sin37^\circ=12\implies pq=40$. 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.

You will get only one unique line passing through $D\displaystyle(0,\sqrt 6)$ and $\displaystyle C\left(\frac{24+4\sqrt 6}{5},\frac{12-3\sqrt 6}{5}\right)$ with slope $\displaystyle\frac{10-5\sqrt 6}{10}$.