Let P_1 = (x_1, y_1) and P_2 = (x_2, y_2). The distance between P_1 and P_2 is then $d(P_1, P_2) = \sqrt{(x_2 - x_1)^2 +(y_2 - y_1)^2}$.
Humans as opposed to computers would do well to understand that this is the square root of the sum of squares because of the Pythagorean Theorem; my habit in in-class computations is to write out the radical with the parens and superscripts first and fill in the differences after reminding everybody of this connection. Every now and then somebody gets it.
There’s a “formula” given in the same section for the midpoint of the line segment connecting P_1 to P_2: but I consider it foolishness to think of this as a formula. Average the x co-ordinates for the “new” x; average the y’s for the y.

One immediate consequence of the distance formula (or of the Pythagorean Theorem directly…) is that x^2 + y^2 = 1 is the equation of the unit circle; shifts-and-scalings lead to the circle equation $(x - H)^2 + (y-K)^2 = R^2$
(for the circle with center (H,K) and radius R). One is prepared either to put an appropriate polynomial in x and y into this form (by completing the square) and “read off” the center and radius, or (what is much easier) to write out the appropriate equation for a given center and radius.

The financial section of the exp-and-log chapter is fraught with formulii. And typing ’em out is no walk in the park, as I might have mentioned from time to time. But let’s see. There’s the old $A = P(1+{r\over n})^{nt}$ for compound interest; its limit as n grows without bound gives the continuous compounding formula $A = Pe^{rt}$. Dividing on both sides of these gives the present value formulas (for “P”, when “A” is known); these should not be understood as separate formulas and I’m sure as heck not gonna write ’em out here. I’ll gladly mention that the division I just mentioned is typically denoted in “negative exponent” style. The effective interest formulas are $(1+{r\over n})^n -1$ and $e^r - 1$. As is usual, our hope in presenting this is that having thought it through, you can forget it and remember instead that one is here letting P and t be equal to unity (by selecting units appropriately); the “multiplier” on P, which is one, for one year, minus the original “one” (one hundred percent of P), is the total interest earned during that year (with its n compounding periods). But wait. Say something once; why say it again. This “formula sheet” project isn’t going to be the affair of an hour and I was crazy to think it would be. Anyway, as usual, there are typically real good summaries in our or any such book.

Better still, we’re on the net. Google “precalculus formula sheet” and look around. Finding the right ones from the best few you find sounds like a darn good exercise. I should have started here.