To find the coordinates of point \( P \) where the medians intersect, let's start by taking a systematic approach based on your steps.
Step 1: Drawing the Medians of the Triangle
Assume we have a triangle defined by vertices:
- A at \( (0, 0) \)
- B at \( (a, 0) \)
- C at \( (b, c) \)
The midpoints of each side are:
- Midpoint \( D \) between B and C: \[ D = \left(\frac{a + b}{2}, \frac{c}{2}\right) \]
- Midpoint \( E \) between A and C: \[ E = \left(\frac{b}{2}, \frac{c}{2}\right) \]
- Midpoint \( F \) between A and B: \[ F = \left(\frac{a}{2}, 0\right) \]
The medians are drawn from each vertex to the midpoint of the opposite side:
- From A (0,0) to D \(\left(\frac{a + b}{2}, \frac{c}{2}\right)\)
- From B (a,0) to E \(\left(\frac{b}{2}, \frac{c}{2}\right)\)
- From C (b,c) to F \(\left(\frac{a}{2}, 0\right)\)
Step 2: Finding the Midpoints
You’ve already defined the midpoints correctly.
Step 3: Find the Coordinates of P, the Centroid
The centroid \( P \) of a triangle is found by averaging the coordinates of the vertices: \[ P = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \] Given \( A(0, 0) \), \( B(a, 0) \), \( C(b, c) \):
- \( P = \left(\frac{0 + a + b}{3}, \frac{0 + 0 + c}{3}\right) = \left(\frac{a + b}{3}, \frac{c}{3}\right) \)
Step 4: Finding Coordinates of P on Each Median
Now let’s find point \( P \) on the median from A to D:
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For the median from A to D: \( P_A = \frac{1}{3} (0, 0) + \frac{2}{3} \left(\frac{a + b}{2}, \frac{c}{2}\right) \): \[ P_A = \left(0 + \frac{2(a + b)}{6}, 0 + \frac{2c}{6}\right) = \left(\frac{a + b}{3}, \frac{c}{3}\right) \]
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For the median from B to E: \( P_B = \frac{1}{3} (a, 0) + \frac{2}{3} \left(\frac{b}{2}, \frac{c}{2}\right) \): \[ P_B = \left(\frac{1}{3}a + \frac{2}{3}\cdot\frac{b}{2}, \frac{1}{3} \cdot 0 + \frac{2}{3} \cdot \frac{c}{2}\right) = \left(\frac{1}{3}a + \frac{b}{3}, \frac{c}{3}\right) = \left(\frac{a + b}{3}, \frac{c}{3}\right) \]
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For the median from C to F: \( P_C = \frac{1}{3} (b, c) + \frac{2}{3} \left(\frac{a}{2}, 0\right) \): \[ P_C = \left(\frac{1}{3}b + \frac{2}{3}\cdot\frac{a}{2}, \frac{1}{3}c + \frac{2}{3}\cdot0\right) = \left(\frac{1}{3}b + \frac{a}{3}, \frac{c}{3}\right) = \left(\frac{a + b}{3}, \frac{c}{3}\right) \]
Conclusion
In all three cases, regardless of which median you analyze, you consistently arrive at the same centroid point: \[ P = \left(\frac{a + b}{3}, \frac{c}{3}\right) \] Thus, the coordinates of \( P \) – the intersection point of the medians – indeed are \( P\left(\frac{a + b}{3}, \frac{c}{3}\right) \).