How can we find the equation of the curve as shown in the figure? 
Above are the illustrations, below in the red box is the QUESTION
'$a$' and '$b$' are the length of the lines as shown in the figure. The two lines are divided into '$n$' equal parts and lines are drawn according to the illustration. The curve formed by the intersection of those lines is the curve with blue stroke. $\theta = \text{Angle between the two lines}$
 A: Given $O(0,0)$, $A(a\cos\theta,a\sin\theta)$,
$B(b,0)$
consider points 
$P_t$, $P_{t+\Delta t}$ on $OA$,
$Q_t$, $Q_{t+\Delta t}$ on $OB$:
\begin{align} 
P_t&=A\,(1-t)
,\\
Q_t&=B\,(t+\Delta t)
,\\
P_{t+\Delta t}&=A\,(1-(t+\Delta t))
,\\
Q_{t+\Delta t}&=B(t+2\Delta t)
.
\end{align}  

Lines $P_tQ_t$, $P_{t+\Delta t}Q_{t+\Delta t}$
intersect at a point $z=(x(t),y(t))$,
\begin{align} 
x(t)&=
(\Delta t+1)^{-1}(2b \Delta t^2
+(a\cos\theta (t-1)+3 t b)\Delta t
\\
&+a\cos\theta-2ta\cos\theta+(b+a\cos\theta)t^2)
,\\
y(t)&=(\Delta t+1)^{-1} 
(t-1+\Delta t)(t-1)a\sin\theta
.
\end{align}
\begin{align}
\lim_{\Delta t \to 0}x(t)
&=
{a\cos\theta-2t a\cos\theta+(b+a\cos\theta)t^2}
\\
&=
a\cos\theta\cdot(1-t)^2+0\cdot2(1-t)t+b\cdot t^2
,\\
\lim_{\Delta t \to 0}y(t)
&=
{(t-1)^2 a\sin\theta}
\\
&=
a\sin\theta\cdot(t-1)^2+0\cdot2(1-t)t+0\cdot t^2
,\\
\\
(x(t),\,y(t))&=A\cdot(1-t)^2+2 O\cdot(1-t)t+B\cdot t^2
,
\end{align}  
which is a well-known quadratic Bezier segment
with control points $A,O,B$.
A: Here's a solution for $a=b=1$ and $\theta=\pi/2$. The result is trivially scaled for any $a$ and $b$, but generalizing for other angles might be more difficult.
First, let $d$ be our step size, so that:
$$d=\frac{1}{n}$$
Then according to the rules above we can see that:
$$x=kd \\ y=1+d-kd$$
Where $k=0,1,\dots,n$.
Let's find the equation of the $k$th straight line, which forms our curve. From two known points $(kd,0)$ and $(0,1+d-kd)$ we obtain the equation:
$$y_k=1+d-kd- \left( \frac{1+d}{kd}-1 \right) x$$
But the straight lines approach the curve only between the intersections with adjacent lines:
$$x_{k1}: \qquad y_{k-1}(x)=y_k(x) \\ x_{k2}: \qquad y_{k+1}(x)=y_k(x)$$
Solving the equations, we take the arithmetic mean as a point which belongs to both the straight line and the curve:
$$x_k=\frac{x_{k1}+x_{k2}}{2}=\frac{k^2 d^2}{1+d}$$
Thus, the equation for our curve at each point is described by:
$$y_k(x_k)=1+d-kd- \left( \frac{1+d}{kd}-1 \right) \frac{k^2 d^2}{1+d}$$
Or, after simplifications:
$$y_k(x_k)=1+d-2kd+\frac{k^2 d^2}{1+d}$$
Or:
$$y_k \left( \frac{k^2 d^2}{1+d} \right)=1+d-2kd+\frac{k^2 d^2}{1+d}$$
Let's introduce a new variable:
$$z=\frac{k^2 d^2}{1+d}$$
Then we have:
$$y \left( z \right)=1+d-2 \sqrt{(1+d)z}+z$$
Now we remember that our step size is supposed to be infinitely small, so we take the limit:
$$\lim_{d \to 0} y(z)=1-2 \sqrt{z}+z$$
Simplifying and renaming our variable we have:

$$y(x)=(1- \sqrt{x})^2$$

Our curve is supposed to be symmetric around $y=x$, which is true in this case.
Here's an illustration for $d=1/10$ compared to $y(x)$:

