Question

Find the co-ordinates of the points of trisection of the line segment joining the points \((-2, 2)\) and \((7, -4)\).

Answer 1

Step 1: Understanding the problem:
We are given the points $A(-2, 2)$ and $B(7, -4)$, and we need to find the coordinates of the points that divide the line segment $AB$ into three equal parts (i.e., the trisection points).
The formula for dividing a line segment in the ratio $m:n$ is:
\[ \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \] where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of the two endpoints.

Step 2: Dividing the line segment into three equal parts:
The points of trisection divide the segment $AB$ in the ratio $1:2$ and $2:1$. So, we will use the section formula to find the coordinates of both trisection points.

Point P dividing AB in the ratio 1:2:
Using the section formula, the coordinates of point $P$ are: \[ P = \left( \frac{1 \cdot 7 + 2 \cdot (-2)}{1 + 2}, \frac{1 \cdot (-4) + 2 \cdot 2}{1 + 2} \right) \] Simplifying the calculations: \[ P = \left( \frac{7 - 4}{3}, \frac{-4 + 4}{3} \right) = \left( \frac{3}{3}, \frac{0}{3} \right) = (1, 0) \] Step 3: Point Q dividing AB in the ratio 2:1:
Similarly, the coordinates of point $Q$ are: \[ Q = \left( \frac{2 \cdot 7 + 1 \cdot (-2)}{2 + 1}, \frac{2 \cdot (-4) + 1 \cdot 2}{2 + 1} \right) \] Simplifying the calculations: \[ Q = \left( \frac{14 - 2}{3}, \frac{-8 + 2}{3} \right) = \left( \frac{12}{3}, \frac{-6}{3} \right) = (4, -2) \] Step 4: Conclusion:
The coordinates of the points of trisection of the line segment joining $A(-2, 2)$ and $B(7, -4)$ are: - Point $P(1, 0)$ (the first trisection point) - Point $Q(4, -2)$ (the second trisection point)