Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are A (4, 7) and B (16, -2).

P divides AB in the ratio 1 : 2 (AP : PB = 1 : 2). Equivalently, P is obtained by moving from A one-third of the way to B:

$$P = A + \frac{1}{3}(B - A)$$

Write $B - A = (16 - 4, -2 - 7) = (12, -9)$.

$$\text{Then } P = \left( 4 + \frac{1}{3}(12) \right), \left( 7 + \frac{1}{3}(-9) \right)$$

$$= (4 + 4, 7 - 3) = (8, 4).$$

Similarly, Q divides AB in the ratio 2 : 1 (AQ : QB = 2 : 1).

Move from A two-thirds of the way to B:

$$Q = A + \frac{2}{3}(B - A) = \left( 4 + \frac{2}{3}(12), 7 + \frac{2}{3}(-9) \right)$$

$$= (4 + 8, 7 - 6) = (12, 1).$$

Midpoint-based verification (connecting with the previous problem):

$$\text{Midpoint of } PQ = \frac{8+12}{2}, \frac{4+1}{2} = (10, 2.5).$$

$$\text{Midpoint of } AB = \frac{4+16}{2}, \frac{7+(-2)}{2} = (10, 2.5).$$

These agree consistent with P, Q being equally spaced about the midpoint.

Length verification:

$$AP^2 = (8 - 4)^2 + (4 - 7)^2 = 16 + 9 = 25,$$

so AP = 5.

$$PQ^2 = (12 - 8)^2 + (1 - 4)^2 = 16 + 9 = 25,$$

so PQ = 5.

$$QB^2 = (16 - 12)^2 + (-2 - 1)^2 = 16 + 9 = 25,$$

so QB = 5.

AP = PQ = QB, confirming the trisection.