Loci of intersection of lines with positive intercepts?? [closed]

These are quadratic Bezier curves obtained by joining regularly spaced points on two line segments.

See paragraph "string art" in

https://en.wikipedia.org/wiki/B%C3%A9zier_curve

The curve is called an envelope.

The envelope of a family of straight lines is the curve, if it exists, having these straight lines as its tangents.

There is a general method that allows to compute the envelope of a family of straight lines (or more generaly a family of curves depending on a parameter $p$). Let us take one of the examples you have displayed in convenient axes i.e., with unit segments $[0,1]$ on x-axis and y-axis, resp.

The equation of the straight line joining point $(p,0)$ to point $(0,1-p)$ is:

$$\dfrac{x}{p}+\dfrac{y}{1-p}=1$$ or, simpler

$$(1-p)x + p y = p(1-p) \ \ $$

(say, for $0 \leq p \leq 1$).

Let us write down the equation obtained by equating the derivatives of the two sides with respect to parameter $p$ (partial derivative, which means that $x$ and $y$ are considered as constants), and group this second equation with the first one:

$$\begin{cases}(1-p)x + p y & = & p(1-p)\\ -x+y & = &1-2p \end{cases}.$$

and now solve this parametric system to get an expression of $x$ and of $y$ as functions of $p$, i.e.,

$$\begin{cases} x & = & p^2\\ y & = & (1-p)^2\end{cases}.$$

which, in fact, describes a parabola with its axis tilted at 45 degrees.

Remark: The method of derivatives is not magic: there is a rather easy proof for it. I present it now.

Edit: Why do we take partial derivative with respect to parameter $p$ ?

Let us use a reasoning that dates back to the 18th century.

We take 2 infinitely close straight lines of the same family, with equations:

$$\left\{\begin{matrix} f(p)x + g(p) y & = & h(p) & \ \ \ \ & (1) \\ f(p+dp)x+g(p+dp)y & = &h(p+dp) & \ \ \ \ & (2) \end{matrix}\right.$$

We consider that their point of intersection is arbitrarily close to the envelope curve, thus in fact is part of the envelope.

This system (1)+(2) is equivalent to the system obtained by keeping (1) and replacing (2) by the difference (2)-(1) (equivalence because one can go way and back from the first system to the second one), that is:

$$\left\{\begin{matrix} f(p)x + g(p) y & = & h(p) & \ \ \ \ & (1) \\ (f(p+dp)-f(p))x+(g(p+dp)-g(p))y & = &h(p+dp)-h(p) & \ \ \ \ & (2') \end{matrix}\right.$$

Dividing now the second equation by $dp$, we obtain the derivatives of coefficients $f(p), g(p), h(p)$ as we have done in the above example.

Remark: the resulting linear system

$$\left\{\begin{matrix} f(p)x + g(p) y & = & h(p) & \ \ \ \ & (1) \\ f'(p)x+g'(p)y & = &h'(p) & \ \ \ \ & (2') \end{matrix}\right.$$

will have solutions for all $p$ such that:

$$\begin{vmatrix} f(p) & g(p) \\ f'(p) & g'(p) \end{vmatrix} \neq 0$$