Is this a coincidence?

Question

I was solving a problem and I noticed something which was very curious, the problem itself was easy:

Find the point of intersection of the lines: $$2x+3y=5$$ and $$(1+c)x+(2+c)y=4+c$$ when c approaches 1.

I solved them and used the limit to get the point as $(-2,3)$ but also notice that for c very near to 1, the second equation is the same as the first! and also one thing that surprised me was that if I chose to put $c=0$, in the second equation, I get it as $$x+2y=4$$ and which remarkably has the same point of intersection with the first line as was when c approached 1. So my questions are

1.)Are these observations mere coincidences?

2.)Is there any geometric significance of these observations?

3.)And lastly, Suppose I had another function of c instead of the ones which I had in the second equation, example $$(1+c^2)x+(4-c^3)y=3+2c^2$$ and I had to find the point of intersection again and use $\lim_{c \to 1}$I don't get the same answer I get when I use $c=0$, (why does it work only for linear functions of c?)

Thanks!

Answer 1

The line $(1+c)x+(2+c)y-4-c=0$ is actually a combination of two lines because. $$(1+c)x+(2+c)y-4-c=(x+2y-4)+\color{red}{c}(x+y-1)=0.$$ Thus for all values of $c$ it represents a line that passes through the intersection of the lines $x+2y-4=0$ and $x+y-1=0$ which is the point $(-2,3)$.

Answer 2

If $2x+3y=5 $ and $(1+c)x+(2+c)y=4+c $, subtracting $\frac{1+c}{2}$ of the first from the second, $4+c-5\frac{1+c}{2} =y((2+c)-3\frac{1+c}{2}) $ or $\frac{2(4+c)-5(1+c)}{2} =y\frac{2(2+c)-3(1+c)}{2} $ or $3-3c =y(1-c) $ or, surprisingly, $y = 3$ for all $c$.

Then $x =\frac{5-3y}{2} =\frac{-4}{2} =-2 $.

Therefore the second line passes through $(-2, 3)$ for all $c$ as does the first.

Note that when $c=1$ the two lines are the same.

Answer 3

From the setup of the question, we have something resembling a system of two simultaneous first-degree equations in $x$ and $y$ (equations of lines in the $x,y$ plane), except that the "constants" of the second equation are all functions of some parameter $c$. Moreover, if $c$ were set to $1$, the two equations would describe the same line.

The question is, what constraints are there on the functions of $c$ if the point of intersection of the two lines as $c$ approaches $1$ must be the same as the point of intersection when $c=0$?

For the case where the functions are first-degree polynomials, the two equations must fit the following form, where $p$, $q$, $r$, $s$, $t$, $u$, and $v$ are all constants of our choosing: \begin{align} px + qy &= r \tag1\\ (sp + t(c - 1))x + (sq + u(c - 1))y &= sr + v(c - 1). \tag2 \end{align}

The reason for writing the equations this way is that equation $(2)$ has to become something of the form $spx + sqy = sr$ when $c=1$, which accounts for the way $p$, $q$, $r$, and $s$ occur in equation $(2)$, while the "first-degree polynomial" allows us to add some multiple of $c - 1$ (and nothing else) to each of the coefficients.

Equation $(1)$ implies that $spx + sqy$ and $sr$ are equal, so subtract this quantity from the left and right sides of equation $(2)$ to produce $$ t(c-1)x + u(c-1)y = v(c-1), $$ so for whenever $c\neq 1$, we can cancel the common factor $c - 1$ to obtain $$ tx + uy = v. \tag3 $$

So the simultaneous equations $(1), (2)$ imply the simultaneous equations $(1), (3)$. If these have a unique solution, that solution is also the solution of the system $(1), (2)$ whenever $c\neq 1$, so it's also the limit of the solution as $c$ approaches $1$ and the value of the solution when $c = 0$. And all of this is dictated by a few constraints we initially put on these equations.

The same solution still works if we replace all the occurrences of $c - 1$ in equation $(2)$ with some other function of $c$, as long as we use the same function in all three places. If we use different functions (not equal, and not constant multiples of each other), then we lose equation $(3)$, and in the general case the solution of the system $(1), (2)$ will vary as we vary $c$.

Note that we don't have to go to higher powers to make this happen: if we change one of the $c-1$ functions to a constant, we can lose the fixed nature of the intersection point. For example, if we make the coefficient of $x$ constant in equation $(1)$, we cannot cancel common factors of $c-1$ and the best we can do for equation $(3)$ is something like $$ tx + u(c-1)y = v(c-1). $$ Except in some special cases, that makes the solution for the intersection point dependent on $c$.