Question

Find the coordinates of the points of trisection of the line segment joining the points A\(-1, 4\) and B\(-3, -2\).

• A. \(\left(-\frac{5}{3}, 2\right)\) and \(\left(-\frac{7}{3}, 0\right)\)
• B. \(\left(-\frac{4}{3}, 2\right)\) and \(\left(-\frac{5}{3}, 1\right)\)
• C. \(\left(-\frac{5}{3}, -2\right)\) and \(\left(-\frac{7}{3}, 1\right)\)
• D. (1, 2) and (2, 0)

Hint:

Once you have calculated the coordinates of the first point \(P\), you can find \(Q\) much faster by recognizing that \(Q\) is simply the midpoint of the line segment \(PB\):
\[ x_Q = \frac{x_P + x_B}{2} = \frac{-\frac{5}{3} - 3}{2} = \frac{-\frac{14}{3}}{2} = -\frac{7}{3} \]
\[ y_Q = \frac{y_P + y_B}{2} = \frac{2 + (-2)}{2} = 0 \]
Using the midpoint formula for the second point is easier and reduces the chance of manual error!

Answer 1

The Correct Option is A

Step 1: Use a vector (step-by-step displacement) approach instead of the section-formula fractions.
The two points of trisection, call them $P$ and $Q$, split segment $AB$ into three equal parts, so moving from $A$ to $P$, from $P$ to $Q$, and from $Q$ to $B$ all cover the same displacement. That means $P$ is one-third of the way from $A$ to $B$, and $Q$ is two-thirds of the way from $A$ to $B$.

Step 2: Find the total displacement vector from $A$ to $B$.
With $A(-1, 4)$ and $B(-3, -2)$:
\[ \vec{AB} = (-3 - (-1),\ -2 - 4) = (-2,\ -6) \]

Step 3: Find $P$ by moving one-third of this displacement from $A$.
\[ P = A + \frac{1}{3}\vec{AB} = \left(-1 + \frac{1}{3}(-2),\ 4 + \frac{1}{3}(-6)\right) \]
\[ P = \left(-1 - \frac{2}{3},\ 4 - 2\right) = \left(-\frac{5}{3},\ 2\right) \]

Step 4: Find $Q$ by moving two-thirds of this displacement from $A$.
\[ Q = A + \frac{2}{3}\vec{AB} = \left(-1 + \frac{2}{3}(-2),\ 4 + \frac{2}{3}(-6)\right) \]
\[ Q = \left(-1 - \frac{4}{3},\ 4 - 4\right) = \left(-\frac{7}{3},\ 0\right) \]

Step 5: Final Answer.
The points of trisection are $\left(-\frac{5}{3}, 2\right)$ and $\left(-\frac{7}{3}, 0\right)$, so option (A) is correct. \[ \boxed{\left(-\dfrac{5}{3}, 2\right) \text{ and } \left(-\dfrac{7}{3}, 0\right)} \]