 # When Nine Points Are Worth But Eight: Euler’s Resolution of Cramer’s Paradox - Three or Four in a Row

Author(s):

Euler also considered the possibility of only four points in a line. For example, given the points $(0,0),$ $(1,1),$ $(2,2),$ $(3,3)$ and $(4,4),$ we might replace $(4,4)$ with some point $A$ not on the line $y=x.$ Let's suppose that $A$ is $(1,2),$ for example, and substitute those values into the equation $(y-x)(\alpha y - \gamma x + \delta) = 0$ from the preceding section. This will give us $2\alpha - \gamma + \delta = 0.$ If we solve this equation for $\delta$ and substitute it into $\alpha y - \gamma x + \delta=0,$ we get $y = mx + (2-m), \quad \mbox{where} \quad m = \frac{\gamma}{\alpha}.$

This is the equation of a line with arbitrary slope $m$ and intercept $b=2-m,$ so that it passes through the point $(1,2).$ As Euler said, there is "one coefficient to be determined," which we can think of as the slope of the line. We observe that we could also have determined this equation by substituting the point $(1,2)$ into the equation $(y-x)(y - mx - b) = 0.$

Finally, let's consider a case that Euler didn't mention but was certainly familiar with:  the case of three points in a line. In this case, the coefficients are all determined, but the graph of the equation is still exceptional.

Figure 7.  A degenerate, two-line conic section. Move points $A$ and $B$ to explore the possibilities. (Interactive applet created using GeoGebra.)

Given the points $(0,0),$ $(1,1),$ $(2,2),$ $(3,3)$ and $(1,2),$ we replace the point $(3,3)$ with another point $B$ not on the line $y=x.$  Let's use the point $(2,1)$ as an example; in the applet in Figure 7 above, $B$ can be any point at all.  If we substitute $(2,1)$ into the equation $(y-x)(y - mx - b) = 0,$ we get $b=1-2m.$ But because we already know that that $b=2-m,$ we have $m=-1,$ which is the slope of the line passing through $(1,2)$ and $(2,1).$ This means that the equation $(y-x)(y - mx - b) = 0$ becomes $(y−x)(y+x−3)=0,$ whose graph is the union of the graphs of the lines $y=x$ and $y=-x+3.$

This situation is an important case of the degenerate conic. Whenever we have three collinear points and two other points not on that line, then the equation of the conic is always uniquely determined, up to a multiplicative factor, but it is an equation that can be factored into a product of two linear equations, one of which is satisfied by the three collinear points, the other of which is satisfied by the two additional points. Probably the most familiar example of a degenerate conic has the equation $y^2-x^2=0,$ which factors as $(y-x)(y+x)=0.$ However, if four or five points are collinear, then the equation has an undetermined linear factor and there are infinitely many conic sections passing through the given points.

Robert E. Bradley (Adelphi University) and Lee Stemkoski (Adelphi University), "When Nine Points Are Worth But Eight: Euler’s Resolution of Cramer’s Paradox - Three or Four in a Row," Convergence (February 2014)

## Dummy View - NOT TO BE DELETED

• • • 