# Collinear points

Alignments to Content Standards: A-REI.D.10

Consider three points in the plane, $P=(-4, 0), Q=(-1, 12)$ and $R=(4, 32)$.

1. Find the equation of the line through $P$ and $Q$.
2. Use your equation in (a) to show that $R$ is on the same line as $P$ and $Q$.
3. Show that $P, Q$ and $R$ are on the graph of the equation $y = x^3 + x^2 - 12 x$.
4. Is it possible for $P, Q$ and $R$ to all lie on a parabola of the form $y=ax^2+bx+c$?

## IM Commentary

This task leads students through a series of problems which illustrate a crucial interplay between algebra (e.g., being solutions to equations) and geometry (e.g., being points on a curve). Students first confirm that $P, Q$ and $R$ are collinear, and then explore other types of curves on which all three points could reside. In particular, they conclude that there does exist such a cubic curve, but no such parabola exists.

When factored, the cubic equation is $y = x(x-3)(x+4)$. It is easy to graph by hand since it has roots at $x=0, 3$ and $-4$ and also goes through $P, Q$ and $R$.

Part (d) is an exploration with multiple possible approaches. They might arrive at a negative answer geometrically, by noting that a secant to a parabola crosses the parabola at most twice. Alternatively, students might try an algebraic approach, discovering that all three points can satisfy the equation $y = a x^2 + b x + c$, only if $a=0$, recovering the original line from part (a).

Instructors might profitably use this task to instigate a discussion about the relationship between algebraic and geometric approaches, perhaps having students share methods of each type.

## Solution

1. Students may try to use slope-intercept form or slope-point form for the equation of a line. Since two points are given, they will find the slope first as rise-over-run or $$m=(y_2-y_1)/(x_2-x_1) = (12-0)/(-1-(-4)) = 12/3 =4.$$

If they use slope-intercept form, they will then plug one point into the equation, $y=4x+b$ and find $b=16$, so $y=4x+16$.

If they use slope-point form, they will plug a point into $y-y_1=4(x-x_1)$ to get either of the equivalent forms $y=4(x+4)$ or $y-12=4(x+1)$.

2. To see that $R$ is on the same line, students decide if the point $(x,y)=(4,32)$ satisfies their equation from the previous part. For example, given $y=4x+16$, students can verify that $32=4(4)+16$. We conclude that $R$ is on the same line as $P$ and $Q$.

Other solution approaches abound: Students might, for example, also check that the slope between $R$ and either $P$ or $Q$ is 4, so they must lie on the same line.

3. Students will most easily check this by confirming that the equation $y=x^3+x^3-12x$ is true after substituting in the $x$-coordinates and $y$-coordinates of each of $P$, $Q$ and $R$.
4. We can argue geometrically as follows: A straight line crosses a parabola at most twice, so since $P$, $Q$, and $R$ are colinear, it cannot be that all three of them are on any parabola. (Note that this answer requires the convention that lines do not count as parabolas. Without this convention, the "degenerate parabola" consisting of the line $y=4x+16$ itself does contain all three points.)

Algebraically, we can proceed by solving the system of equations we get by evaluating the expression $y=ax^2+bx+c$ at the $x$- and $y$-coordinates of $P$, $Q$, and $R$. This gives \begin{eqnarray*} 0&=&a (-4)^2+b(-4)+c\\ 12&=&a(-1)^2+b(-1)+c\\ 32&=&a(4)^3+b(4)+c \end{eqnarray*} Solving the resulting system gives $a=0, b=4$ and $c=16$.

It may be worth remarking that, in this case, the geometric approach proves a more general statement, as it applies to all parabolas, and not just those of the form $y=ax^2+bx+c$.