4.1 Q-13

Question Statement

Find the points that trisect the line segment joining the points $A(-1, 4)$ and $B(6, 2)$.

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Background and Explanation

In this problem, we are asked to find the points that trisect the line segment joining two given points. Trisection means dividing the line segment into three equal parts, which results in two points that divide the segment in the ratio 1:2 and 2:1. We will use the section formula, which is used to find the coordinates of a point dividing a line segment in a given ratio.

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Solution

We are given points $A(-1, 4)$ and $B(6, 2)$. We need to find two points $C$ and $D$ that trisect the line segment $AB$.

Step 1: Use the Section Formula for Point $C$

Point $C$ divides the segment in the ratio 1:2. According to the section formula, the coordinates of a point dividing a line segment in the ratio $m:n$ are given by:

$$x = \frac{n x_1 + m x_2}{m + n}, \quad y = \frac{n y_1 + m y_2}{m + n}$$

For point $C$, we use the ratio 1:2:

$$x_1 = -1, , y_1 = 4 \quad \text{(coordinates of $A$)}$$ $$x_2 = 6, , y_2 = 2 \quad \text{(coordinates of $B$)}$$

Using the section formula for $C(x_1, y_1)$ where the ratio is 1:2:

• For the $x$-coordinate:

$$x_1 = \frac{1(6) + 2(-1)}{1 + 2} = \frac{6 - 2}{3} = \frac{4}{3}$$

• For the $y$-coordinate:

$$y_1 = \frac{1(2) + 2(4)}{1 + 2} = \frac{2 + 8}{3} = \frac{10}{3}$$

Thus, the coordinates of point $C$ are $\left(\frac{4}{3}, \frac{10}{3}\right)$.

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Step 2: Use the Section Formula for Point $D$

Point $D$ divides the segment in the ratio 2:1. We again use the section formula but with the ratio 2:1:

• For the $x$-coordinate:

$$x_2 = \frac{2(6) + 1(-1)}{2 + 1} = \frac{12 - 1}{3} = \frac{11}{3}$$

• For the $y$-coordinate:

$$y_2 = \frac{2(2) + 1(4)}{2 + 1} = \frac{4 + 4}{3} = \frac{8}{3}$$

Thus, the coordinates of point $D$ are $\left(\frac{11}{3}, \frac{8}{3}\right)$.

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Step 3: Verify by Considering Midpoint

Alternatively, point $D$ is also the midpoint of segment $CB$. We can calculate the midpoint of $CB$ to confirm the result.

• The formula for the midpoint of a segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is:

$$x_{\text{mid}} = \frac{x_1 + x_2}{2}, \quad y_{\text{mid}} = \frac{y_1 + y_2}{2}$$

Using the coordinates of $C\left(\frac{4}{3}, \frac{10}{3}\right)$ and $B(6, 2)$:

• For the $x$-coordinate:

$$x_2 = \frac{\frac{4}{3} + 6}{2} = \frac{\frac{4}{3} + \frac{18}{3}}{2} = \frac{22}{6} = \frac{11}{3}$$

• For the $y$-coordinate:

$$y_2 = \frac{\frac{10}{3} + 2}{2} = \frac{\frac{10}{3} + \frac{6}{3}}{2} = \frac{16}{6} = \frac{8}{3}$$

This confirms that the coordinates of point $D$ are indeed $\left(\frac{11}{3}, \frac{8}{3}\right)$.

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Key Formulas or Methods Used

• Section Formula:
  For a point dividing a line segment in the ratio $m:n$:

$$x = \frac{n x_1 + m x_2}{m + n}, \quad y = \frac{n y_1 + m y_2}{m + n}$$

• Midpoint Formula:
  The midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is:

$$x_{\text{mid}} = \frac{x_1 + x_2}{2}, \quad y_{\text{mid}} = \frac{y_1 + y_2}{2}$$

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Summary of Steps

1. Use the section formula to find the first trisection point $C$ (ratio 1:2).
2. Use the section formula to find the second trisection point $D$ (ratio 2:1).
3. Verify the coordinates of $D$ by using the midpoint formula between $C$ and $B$.