Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Step-by-Step Solution
Key Concept: Use the section formula (or vector method) for internal division of a line segment. For trisection, the segment is divided in the ratios 1:2 and 2:1.
1. Let the end points be \(A(4,-1)\) and \(B(-2,-3)\).\
2. The vector \(\overrightarrow{AB}\) = \((-2-4,\; -3-(-1))\) = \((-6,\; -2)\).\
3. For the first point of trisection \(P\) (closer to \(A\)), the ratio \(AP:PB = 1:2\). Hence \(P\) is located at one‑third of the way from \(A\) to \(B\):\
\[\begin{aligned}
x_P &= x_A + \frac{1}{3}(x_B - x_A) = 4 + \frac{1}{3}(-2-4) = 4 - 2 = 2,\\
y_P &= y_A + \frac{1}{3}(y_B - y_A) = -1 + \frac{1}{3}(-3+1) = -1 - \frac{2}{3} = -\frac{5}{3}.
\end{aligned}\]
So \(P(2, -\frac{5}{3})\).\
4. For the second point of trisection \(Q\) (closer to \(B\)), the ratio \(AQ:QB = 2:1\). Hence \(Q\) is located at two‑thirds of the way from \(A\) to \(B\):\
\[\begin{aligned}
x_Q &= x_A + \frac{2}{3}(x_B - x_A) = 4 + \frac{2}{3}(-6) = 4 - 4 = 0,\\
y_Q &= y_A + \frac{2}{3}(y_B - y_A) = -1 + \frac{2}{3}(-2) = -1 - \frac{4}{3} = -\frac{7}{3}.
\end{aligned}\]
So \(Q(0, -\frac{7}{3})\).\
5. Hence the two points that trisect the segment \(AB\) are \((2, -\frac{5}{3})\) and \((0, -\frac{7}{3})\).
Correct Answer: (2, -5/3) and (0, -7/3)