To find the median of a triangle, we need to find the midpoint of each side and then find the line segment connecting each vertex to the midpoint of the opposite side. The triangle vertices are given as:
- \( A(4, 5) \)
- \( B(1, 2) \)
- \( C(6, 2) \)
Step 1: Find the midpoints of each side.
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Midpoint of side BC: \[ M_{BC} = \left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2} \right) = \left( \frac{1 + 6}{2}, \frac{2 + 2}{2} \right) = \left( \frac{7}{2}, 2 \right) = (3.5, 2) \]
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Midpoint of side AC: \[ M_{AC} = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) = \left( \frac{4 + 6}{2}, \frac{5 + 2}{2} \right) = \left( 5, 3.5 \right) \]
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Midpoint of side AB: \[ M_{AB} = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right) = \left( \frac{4 + 1}{2}, \frac{5 + 2}{2} \right) = \left( 2.5, 3.5 \right) \]
Step 2: Define the medians.
Now, we will find the equations of the medians.
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Median from A to M_{BC}(3.5, 2): The slope of the line through points \( A(4, 5) \) and \( M_{BC}(3.5, 2) \): \[ \text{slope} = \frac{2 - 5}{3.5 - 4} = \frac{-3}{-0.5} = 6 \] The equation of the line in point-slope form becomes: \[ y - 5 = 6(x - 4) \implies y = 6x - 24 + 5 \implies y = 6x - 19 \]
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Median from B to M_{AC}(5, 3.5): The slope of the line through points \( B(1, 2) \) and \( M_{AC}(5, 3.5) \): \[ \text{slope} = \frac{3.5 - 2}{5 - 1} = \frac{1.5}{4} = \frac{3}{8} \] The equation of the line in point-slope form is: \[ y - 2 = \frac{3}{8}(x - 1) \implies y - 2 = \frac{3}{8}x - \frac{3}{8} \implies y = \frac{3}{8}x + \frac{13}{8} \]
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Median from C to M_{AB}(2.5, 3.5): The slope of the line through points \( C(6, 2) \) and \( M_{AB}(2.5, 3.5) \): \[ \text{slope} = \frac{3.5 - 2}{2.5 - 6} = \frac{1.5}{-3.5} = -\frac{3}{7} \] The equation of the line in point-slope form is: \[ y - 2 = -\frac{3}{7}(x - 6) \implies y - 2 = -\frac{3}{7}x + \frac{18}{7} \implies y = -\frac{3}{7}x + \frac{32}{7} \]
Conclusion: Finding the point of concurrency.
To show that the medians intersect at the centroid, we can calculate the centroid \((G)\) of triangle ABC. The centroid is given by the average of the vertices' coordinates:
\[ G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right) = \left( \frac{4 + 1 + 6}{3}, \frac{5 + 2 + 2}{3} \right) = \left( \frac{11}{3}, \frac{9}{3} \right) = \left( \frac{11}{3}, 3 \right) \]
Summary of Ordered Pairs for Median Midpoints:
The ordered pairs of the midpoints for each median are:
- \( M_{BC} = (3.5, 2) \)
- \( M_{AC} = (5, 3.5) \)
- \( M_{AB} = (2.5, 3.5) \)
Hence, the centroid of the triangle, which lies on all three medians, is the point \( \left( \frac{11}{3}, 3 \right) \).
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