We have two cases.
i) Think of the circle with center in $A$ and radius $AB$. Any point will do.
or
ii) Get the circle with center in $B$ and radius $AB$. Pick any point.
Remember that the equation of a circle is given by $$(x - x_c)^2 + (y - y_c)^2 = r²$$
where $(x_c, y_c)$ is the center, and $r$ is the radius. Remember also, that $$d((x,y),(a,b)) = \sqrt{(x-a)^2 + (y-b)^2}$$
Can you do it?
Rearranging the equations,
$$L_1:y=\frac12x+2$$
$$L_2:y=-2x+5$$
As product of slope of $L_1$ and $L_2$ $=(-2)\cdot(\frac12)=-1$,
$L_1$ is perpendicular to $L_2$.
Let length of one of the side of triangle be $r$.
$$\frac12r^2=10$$ $$r=\sqrt{20}$$
Let a point on $L_1$ be A: $(t,\frac{t+4}2)$, by considering the distance between the intersection point $(\frac65,\frac{13}5)$ and A,we have
$$\sqrt{20}=\sqrt{(t-\frac65)^2+(\frac{t+4}2-\frac{13}5)^2}$$
$$t=5.2\text{ or }-2.8$$
$$A_1= (5.2,4.6)\text{ or }A_2=(-2.8,0.6)$$
Similarly, for a point on $L_2$, B: $(p,-2p+5)$, by considering the distance between the intersection point $(\frac65,\frac{13}5)$ and B,we have
$$\sqrt{20}=\sqrt{(p-\frac65)^2+(-2p+5-\frac{13}5)^2}$$
$$p=3.2\text{ or }-0.8$$
$$B_1= (3.2,-1.4)\text{ or }B_2=(-0.8,6.6)$$
So we have line $L_{11}=A_1B_1, L_{21}=A_2B_1,L_{22}=A_2B_2,L_{12}=A_1B_2$
$$L_{11}:\frac{y-4.6}{x-5.2}=\frac{4.6-(-1.4)}{5.-3.2}$$
$$L_{11}:y=3x-11$$
Similar for $L_{21},L_{22}\text{ and }L_{12}$.
Best Answer
In fact, there a way to obtain the coordinates of point $A$ without solving a quadratic equation. See program below (sorry, in Matlab, but so cousin to Python...).
Compute vector $\vec{DB}=\frac12 \vec{CB}=\pmatrix{r_1\\s_1}=\pmatrix{\tfrac12(x_B-x_C)\\ \tfrac12(y_B-y_C)}$ and its norm.
Deduce from this norm the length $\ell$ of altitude $AD$ by using Pythagoras in right triangle $ADB$.
Then, due to vector equation :
$$\vec{DA}=\ell \vec{V} \ \ \iff \ \ A=D+\ell \vec{V}$$
the coordinates of $A$ are :
$$\begin{cases}x_A&=&x_D+ \ell r\\ y_A&=&y_D+ \ell s\end{cases}$$
where midpoint $D$ has coordinates $$D=\pmatrix{\tfrac12(x_B+x_C)\\ \tfrac12(y_B+y_C)} $$
and $V=\pmatrix{r\\s}$ is defined in two steps :
(please note that $\vec{W}$ is an orthogonal vector to vector $\vec{DB}$)
Matlab program: