What do you get when you cross the curves r1(t) = t, 3 − t, 48 + t2 and r2(s) = 8 − s, s − 5, s2? The answer is (8/3)(1 – 1/4), which can be simplified to 2. In this blog post we will explore what these three curves have in common and how they are related. The first two curves are identical to the second, except that they differ by an x-intercept. Let’s call them rx(t) = t − x and Ṽy(s) = s – y, respectively. They will always intersect at (0, 0), because both have a slope of zero for all values of their independent variables: rx′(t) + λrx″(t) = 0; Ṽy′(s)+λry″(-s)=0 . This is true for any value in both x and s coordinates on these graphs. This means they also cross each other at (- ∞ , -∞). But what about when we add the third curve? Let’s call it rz(t) = t − z. This is what happens when we subtract λ from both of these curves: Ṽx′(-s)+λrx″(-s)=0; y′(s−λry″= 0 . We can see that this will always be true and they are intersections at (-∞, -∞). This means all three graphs intersect each other in a straight line. It also follows that there cannot be any more than two lines of intersection between them because if you had a fourth equation for another graph it would have to cross one or more values already labeled as an intersection on the first three equations, which isn’t possible