Question

In clip 2, session 2 of part A in the first unit (Harder Problem: Triangles Under the Graph
of y=1/x) can somebody please explain why the answer is simply "two" at the end? He says that the answer is 2x0y0 then suddenly resolves that statement and says it's just two, and I don't understand why.

Answer 1

y=1/x, and y0=1/x0. 2x0y0=2x0(1/x0)=2

Answer 2

This is exactly an example of what Prof. Jerison just said about variables doing double (and sometimes triple duty, etc.)

The point slope form of the line was used to find the y-intercept which turned out to be (2x_0, 0) and the x-intercept (0, 2y_0). Since we are doing triangles against the axes, 2x_0 is also the distance (length) to the right angle at (0,0).

In that vein the area of the triangle is then 1/2 (2x_0) (2y_0) = 2(x_0)(y_0)
and you got that far.

But x_0 and y_0 are also the coordinates of a point on y=1/x. which means it must be the case that x_0 = 1/y_0 or y_0= 1/x_0 if you look at it that way. Whichever way you look at it, (x_0)(y_0) = 1. The function y=1/x can also be written xy = 1.

So

2(x_0)(y_0) = 2(1) = 2

Answer 3

Thank you so much for the very clear and concise explanation, cleared up my question very quickly. Probably should have payed more attention, but thank you regardless!