$$x_3$$ is the $$753^{\text{rd}}$$ AM of 2007 AM's inserted between $$x_1$$ and $$x_2$$. $$y_3$$ is the $$753^{\text{rd}}$$ AM of 2007 AM's inserted between $$y_1$$ and $$y_2$$. Show that $$A(x_1, y_1)$$, $$P(x_3, y_3)$$, $$B(x_2, y_2)$$ are collinear. Determine also the ratio $$AP \colon PB$$.

PLACEHOLDER ANSWER - the answer is a proof and a ratio. Inserting 2007 AMs divides the interval into 2008 equal steps, so $$x_3 = x_1 + \frac{753}{2008}(x_2 - x_1)$$ and $$y_3 = y_1 + \frac{753}{2008}(y_2 - y_1)$$. Hence $$P = A + \frac{753}{2008}(B - A)$$, which lies on segment AB, so the points are collinear. P divides AB in the ratio $$753 \colon 1255 = 3 \colon 5$$.