Consider the figure below:
Let's calculate $CD$ using the Law of cosines in $\triangle CAD$:
$$ CD^2=10^2+200^2-2 \cdot 10 \cdot 200 \cdot \cos 82^{\circ} \Rightarrow$$
$$\Rightarrow CD \approx 198.855 \, \mathrm{m}.$$
Now using the Law of cosines in $\triangle ACD$, let's calculate $\cos \alpha$:
$$ 200^2=10^2+CD^2-2 \cdot 10 \cdot CD \cdot \cos \alpha \Rightarrow$$
$$\Rightarrow \cos \alpha \approx -0.0896857.$$
Note that $\triangle CEF$ is a right-angled triangle, $CE=r$, and $CF=\frac{CD}{2}$, so:
$$\cos(180^{\circ} - \alpha)= \frac{CF}{CE} \Rightarrow$$
$$\Rightarrow -\cos \alpha =\frac{CD}{2r} \Rightarrow$$
$$\Rightarrow r = \frac{CD}{-2\cos \alpha} \Rightarrow$$
$$\Rightarrow r \approx 1108.6 \, \mathrm{m}.$$
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
Let's call your two equations $f(x)=Ax+B$ (the upper one) and $g(x)=Cx+D$.
For exactly the right $x=x_h$ they are $h$ apart. We then have $$f(x_h)-h=g(x_h)$$ from which $x_h$ can be isolated.
At their intercept, with $x=x_I$ we have $f(x_I)=g(x_I)$, from which we can isolate $x_I$.
Then $$x_I-x_h=w$$