Given any three non colinear points of the Cartesian plane:
$$P_1(x_1,y_1)$$ $$P_2(x_2,y_2)$$ $$P_3(x_3,y_3)$$
such that $x_1<x_2<x_3$, there exists an absolute value function:
$$f(x)=a|x-h|+k$$
that goes through all three points, where $\pm a$ is the slope of the two lines ('arms') forming the graph of $f$ and $(h,k)$ is the point where these lines intersect.
Since I have yet to find a counter-example, I would like to prove that the value of $a$ will actually be the fastest slope, among the following:
$$\text{Slope of }{P_1P_2}$$ $$\text{Slope of }{P_2P_3}$$
(For instance, if the slope of $P_1P_2$ is $3$ and that of $P_2P_3$ is $-4$, then $a=-4$).
I am not satisfied with what I came up with and would like some insight in how I could make this more rigorous.
My attempt
The trivial case of having 3 points lie on the same arm of the graph of $f$ is excluded because $P_1$, $P_2$ and $P_3$ are not colinear.
Since the graph of $f$ only has two arms, one arm must support two of the three points and the other arm supports the other point (unless $(h,k)$ is one of the three points, in which case both arms contain 2 of the three points).
Therefore, the slope of one of the arms will have to be one of the following:
$$\text{Slope of }{P_1P_2}$$ $$\text{Slope of }{P_1P_3}$$ $$\text{Slope of }{P_2P_3}$$
However, chosing $P_1P_3$ as one of the arms (or a segment supported by it) is not acceptable because if $P_2$ does not lie on $P_1P_3$ (and it doesn't because these points are not colinear), then $f(x_2)$ is not well-defined (it is either $y_2$, or the value of the line going through $P_1P_3$ above $x_2$). In other words, $f$ would not be defined at $x=x_2$.
So we are left with having to choose the value of $a$ to be among:
$$\text{Slope of }{P_1P_2}$$ $$\text{Slope of }{P_2P_3}$$
This is the part I'm not happy about
- I show that in some given, specific case, taking the slowest slope does not yield the correct absolute value function
- But since there must be an absolute value function that goes through the three points, the case where the slope is fastest must be the right one
For instance, if $P_1=(0,1)$, $P_2=(2,3)$ and $P_3=(4,7)$, then we have:
$$\text{Slope of }P_1P_2 = \frac{3-1}{2-0}=1$$ $$\text{Slope of }P_2P_3 = \frac{7-3}{4-2}=2$$
Chosing $P_1P_2$ to be one of the arms (associated with a slope of $a=1$) means that that arm's equation is: $$y=x+1$$ The other arm's slope must be $-a=-1$ and this other arm must pass through $P_3$, yielding the equation: $$y=-x+11$$ Solving this linear system yields the intersection point $(h,k)=(5,6)$ and thus:
$$f(x)=|x-5|+6$$
However:
$$f(0)=|0-5|+6=11\neq 1$$
So this absolute value function does not go through $P_1$ and therefore isn't the one we were looking for.
Had I taken $a=2$ (the fastest slope), then I would have found, in a similar way, that $(h,k)=(0.5, 0)$ and then:
$$f(x)=2|x-0.5|$$
which verifies all three points.
What I'm not liking about this is that I'm not really showing that the fastest slope was the right choice, but that one slope was the right choice over another. How can I show that the fastest slope will always be the right choice?