Find the co-ordinates of the points of trisection of the straight line joining…

For a point dividing the segment from A to B in the ratio m : n (measured from A), the section formula gives P = A + (m / (m+n)) · (B − A). A line segment's two trisection points split it into three equal parts, corresponding to ratios 1 : 2 and 2 : 1 from A — i.e., P₁ = A + (1/3)(B − A) and P₂ = A + (2/3)(B − A).

1. Compute B − A = (−3 − 1, 4 − (−2)) = (−4, 6).

2. First trisection point: P₁ = A + (1/3)(B − A) = (1 + (1/3)(−4), −2 + (1/3)(6)) = (1 − 4/3, −2 + 2) = (−1/3, 0).

3. Second trisection point: P₂ = A + (2/3)(B − A) = (1 + (2/3)(−4), −2 + (2/3)(6)) = (1 − 8/3, −2 + 4) = (−5/3, 2).

Independent check: for any pair of trisection points, P₁ + P₂ = A + B (the two points are symmetric about the segment's midpoint pair-wise). Here A + B = (1 + (−3), −2 + 4) = (−2, 2), and P₁ + P₂ = (−1/3 + (−5/3), 0 + 2) = (−2, 2) — the computation checks out.

Answer: The points of trisection are (−1/3, 0) and (−5/3, 2).