Loci of intersection of lines with positive intercepts??

Question

How does one go about proving the equation of the curve that is formed by tracing the intersection of the lines with each other?

Answer 1

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.

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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:

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

Answer 2

If two straight lines at an inclination $\alpha $ are each divided into $n$ equal parts from points $P,Q$ on them to their intersection point $O$ and serially connected by straight lines in reverse order, then they include between them a parabolic envelope (touching locus) with $P,Q$ as points of tangentcy.

Example: (when $ \alpha= \pi/2 )$:

Points on $x-,y-$ axes and sub-divisions:

$$ P= (0,b) ; Q ( a,0); p (0,b\, t) ; q ( a(1-t),0); $$

Variable line connecting points$(p,q)$

$$ \frac{x}{a(1-t)} + \frac {y}{b \,t} = 1 \tag{1}$$

To eliminate $t$ by $C-$ discriminant method, partially differentiate w.r.t. $t$ and eliminate from (1),

$$ \frac{x}{a(1-t)^2} = \frac {y}{b \,t^2} \tag{2}$$

Eliminating $t$ between (1) and (2), we have the parabola tangential to axes at the intercepts.

$$ \sqrt{\frac{x}{a}} + \sqrt{\frac{y}{b}} = 1. \tag{3} $$

which are typical of the two graphs at left.