# When Nine Points Are Worth But Eight: Euler’s Resolution of Cramer’s Paradox - Special Case: the Parabola

Author(s):
Robert E. Bradley (Adelphi University) and Lee Stemkoski (Adelphi University)

Given three distinct points $$(x_1, y_1), (x_2, y_2),$$ and $$(x_3, y_3),$$ we wish to determine the conditions under which there exists a parabola that passes through these points. Specifically, let's consider the case in which the parabola gives $$y$$ as a function of $$x.$$ In other words, we want to find real numbers $$a, b, c$$ such that the equation of the parabola $$y = f(x) = ax^2 + bx + c$$ satisfies $$y_i = f(x_i)$$ for $$i = 1, 2, 3.$$

First, we observe that for the function $$f$$ to exist, the values $$x_i$$ must all be distinct, for otherwise we would be assigning two distinct codomain values to a single domain value of $$f.$$

The real numbers $$a, b, c$$ exist exactly when there is a solution to the system of equations

$\begin{eqnarray*}a(x_{1})^{2} + bx_{1} + c & = & y_1 \\a(x_{2})^{2} + bx_{2} + c & = & y_2 \\a(x_{3})^{2} + bx_{3} + c & = & y_3\end{eqnarray*}$

Because this system of equations is linear in $$a, b, c,$$ we may rewrite it as the matrix equation $\begin{bmatrix}x_1^2 & x_1 & 1 \\x_2^2 & x_2 & 1 \\x_3^2 & x_3 & 1\end{bmatrix}\,\begin{bmatrix}a \\b \\c \end{bmatrix}=\begin{bmatrix}y_1 \\y_2 \\y_3\end{bmatrix},$

which may be represented by the augmented matrix $\begin{bmatrix}x_1^2 & x_1 & 1 & y_1 \\x_2^2 & x_2 & 1 & y_2 \\x_3^2 & x_3 & 1 & y_3\end{bmatrix}.$

We now perform some elementary row operations – subtract the first row from the second row and from the third row: $\begin{bmatrix}x_1^2 & x_1 & 1 & y_1 \\x_2^2 - x_1^2 & x_2 - x_1 & 0 & y_2 - y_1 \\x_3^2 - x_1^2 & x_3 - x_1 & 0 & y_3 - y_1 \\\end{bmatrix}$

Because we know that $$x_1\neq x_2$$ and $$x_1 \neq x_3,$$ we may divide the second and third rows by $$x_2 - x_1$$ and $$x_3 - x_1,$$ respectively.  This gives$\begin{bmatrix}x_1^2 & x_1 & 1 & y_1 \\x_2 + x_1 & 1 & 0 & {\displaystyle \frac{y_2 - y_1}{x_2 - x_1}} \\x_3 + x_1 & 1 & 0 & {\displaystyle \frac{y_3 - y_1}{x_3 - x_1}} \\\end{bmatrix}$

Finally, we subtract the second row from the third row, and obtain:$\begin{bmatrix}x_1^2 & x_1 & 1 & y_1 \\x_2 + x_1 & 1 & 0 & {\displaystyle \frac{y_2 - y_1}{x_2 - x_1}} \\x_3 - x_2 & 0 & 0 &{\displaystyle \frac{y_3 - y_1}{x_3 - x_1} - \frac{y_2 - y_1}{x_2 - x_1}} \\\end{bmatrix}$

Because $$x_2\neq x_3,$$ we see that the system of equations represented by this augmented matrix is consistent, and moreover, that the solution for $$a, b, c$$ is unique. Therefore, there is a unique curve with equation $$y = ax^2 + bx + c$$ that passes through the given points.

But is this curve really a parabola?  Only if $$a\ne 0$$; otherwise it's just the straight line $$y=bx+c.$$ Now, the last line of the augmented matrix is equivalent to the relation $a =\frac{m'-m}{x_3-x_2},$ where $$m$$ is the slope of the line joining $$(x_1,y_1)$$ to $$(x_2,y_2)$$ and $$m'$$ is the slope of the line joining $$(x_1,y_1)$$ to $$(x_3,y_3).$$ Thus $$a\ne 0$$ precisely in the case where the three points are noncollinear.

Figure 3.  A parabola passing through three points. Move points $$A,B,$$ and $$C$$ to form parabolas. (Interactive applet created using GeoGebra.)

You can use the applet in Figure 3 to explore this construction graphically. Notice that the curve becomes a straight line if you move one of the points so that it is collinear with the others. Also, if you move one of the points so that it has the same $$x$$-coordinate as one of the others, the curve will disappear, indicating that there is no solution to the linear system in this case.

Robert E. Bradley (Adelphi University) and Lee Stemkoski (Adelphi University), "When Nine Points Are Worth But Eight: Euler’s Resolution of Cramer’s Paradox - Special Case: the Parabola," Loci (February 2014)