To find the ordered pairs of the three median bisectors (the points where the medians intersect the opposite side) for a triangle with vertices A(4,5), B(1,2), and C(6,2), we need to first find the midpoints of the sides of the triangle and then use these midpoints to find the equations of the medians.
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Finding the midpoints of the sides:
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Midpoint of BC: \[ M_{BC} = \left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2} \right) = \left( \frac{1 + 6}{2}, \frac{2 + 2}{2} \right) = \left( \frac{7}{2}, 2 \right) = \left( 3.5, 2 \right) \]
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Midpoint of AC: \[ M_{AC} = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) = \left( \frac{4 + 6}{2}, \frac{5 + 2}{2} \right) = \left( 5, \frac{7}{2} \right) = \left( 5, 3.5 \right) \]
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Midpoint of AB: \[ M_{AB} = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right) = \left( \frac{4 + 1}{2}, \frac{5 + 2}{2} \right) = \left( \frac{5}{2}, \frac{7}{2} \right) = \left( 2.5, 3.5 \right) \]
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Finding the median equations:
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Median from A to M_{BC} (from A(4,5) to M_{BC}(3.5, 2)):
- Slope of AM_{BC}: \[ m_{AM_{BC}} = \frac{y_{M_{BC}} - y_A}{x_{M_{BC}} - x_A} = \frac{2 - 5}{3.5 - 4} = \frac{-3}{-0.5} = 6 \]
- Equation of the median using point-slope form \(y - y_1 = m(x - x_1)\): \[ y - 5 = 6(x - 4) \implies y = 6x - 24 + 5 \implies y = 6x - 19 \]
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Median from B to M_{AC} (from B(1,2) to M_{AC}(5, 3.5)):
- Slope of BM_{AC}: \[ m_{BM_{AC}} = \frac{3.5 - 2}{5 - 1} = \frac{1.5}{4} = \frac{3}{8} \]
- Equation: \[ y - 2 = \frac{3}{8}(x - 1) \implies y - 2 = \frac{3}{8}x - \frac{3}{8} \implies y = \frac{3}{8}x + \frac{13}{8} \]
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Median from C to M_{AB} (from C(6,2) to M_{AB}(2.5, 3.5)):
- Slope of CM_{AB}: \[ m_{CM_{AB}} = \frac{3.5 - 2}{2.5 - 6} = \frac{1.5}{-3.5} = -\frac{3}{7} \]
- Equation: \[ y - 2 = -\frac{3}{7}(x - 6) \implies y - 2 = -\frac{3}{7}x + \frac{18}{7} \implies y = -\frac{3}{7}x + \frac{32}{7} \]
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The medians' equations are:
- Median from A: \(y = 6x - 19\)
- Median from B: \(y = \frac{3}{8}x + \frac{13}{8}\)
- Median from C: \(y = -\frac{3}{7}x + \frac{32}{7}\)
Since these medians intersect at a single point, they all meet at the centroid of the triangle.
- Finding the centroid: The centroid (G) of triangle ABC is given by: \[ G = \left(\frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}\right) = \left(\frac{4 + 1 + 6}{3}, \frac{5 + 2 + 2}{3}\right) = \left(\frac{11}{3}, \frac{9}{3}\right) = \left(\frac{11}{3}, 3\right) \approx (3.67, 3). \]
Therefore, the ordered pairs of the three median bisectors are \(M_{BC}(3.5, 2)\), \(M_{AC}(5, 3.5)\), and \(M_{AB}(2.5, 3.5)\), and they all meet at the centroid \(G \left( \frac{11}{3}, 3 \right)\).
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