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: Understanding the Question:
We are given two points \(A(-1, 4)\) and \(B(-3, -2)\). We need to determine the coordinates of the two points of trisection that divide the line segment \(AB\) into three equal parts.

Step 2: Key Formula or Approach:
Let the points of trisection be \(P\) and \(Q\) such that \(AP = PQ = QB\).
1. Point \(P\) divides the line segment \(AB\) internally in the ratio \(1 : 2\).
2. Point \(Q\) divides the line segment \(AB\) internally in the ratio \(2 : 1\) (or is the midpoint of \(PB\)).
We will use the Section Formula:
\[ x = \frac{mx_2 + nx_1}{m + n}, \quad y = \frac{my_2 + ny_1}{m + n} \]

Step 3: Detailed Explanation:
1.

Find the coordinates of point P (ratio 1 : 2):
Here, \(A(x_1, y_1) = (-1, 4)\), \(B(x_2, y_2) = (-3, -2)\), and \(m : n = 1 : 2\).
- Calculate the \(x\)-coordinate of \(P\):
\[ x_P = \frac{1(-3) + 2(-1)}{1 + 2} = \frac{-3 - 2}{3} = -\frac{5}{3} \]
- Calculate the \(y\)-coordinate of \(P\):
\[ y_P = \frac{1(-2) + 2(4)}{1 + 2} = \frac{-2 + 8}{3} = \frac{6}{3} = 2 \]
So, the coordinates of \(P\) are:
\[ P = \left(-\frac{5}{3}, 2\right) \]

2.

Find the coordinates of point Q (ratio 2 : 1):
Here, \(A(x_1, y_1) = (-1, 4)\), \(B(x_2, y_2) = (-3, -2)\), and \(m : n = 2 : 1\).
- Calculate the \(x\)-coordinate of \(Q\):
\[ x_Q = \frac{2(-3) + 1(-1)}{2 + 1} = \frac{-6 - 1}{3} = -\frac{7}{3} \]
- Calculate the \(y\)-coordinate of \(Q\):
\[ y_Q = \frac{2(-2) + 1(4)}{2 + 1} = \frac{-4 + 4}{3} = \frac{0}{3} = 0 \]
So, the coordinates of \(Q\) are:
\[ Q = \left(-\frac{7}{3}, 0\right) \]

Step 4: Final Answer:
The coordinates of the points of trisection are \(\left(-\frac{5}{3}, 2\right)\) and \(\left(-\frac{7}{3}, 0\right)\), which corresponds to option (A).