To find the medians of triangle ABC with vertices A(4,5), B(1,2), and C(6,2), we first need to find the midpoints of each side of the triangle. Then, we can create the equations of the medians that connect each vertex to the midpoint of the opposite side.
-
Finding the midpoints of each side:
-
Midpoint of BC: \[ M_{BC} = \left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2} \right) = \left( \frac{1 + 6}{2}, \frac{2 + 2}{2} \right) = \left( \frac{7}{2}, 2 \right) = \left( 3.5, 2 \right) \]
-
Midpoint of AC: \[ M_{AC} = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) = \left( \frac{4 + 6}{2}, \frac{5 + 2}{2} \right) = \left( \frac{10}{2}, \frac{7}{2} \right) = \left( 5, 3.5 \right) \]
-
Midpoint of AB: \[ M_{AB} = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right) = \left( \frac{4 + 1}{2}, \frac{5 + 2}{2} \right) = \left( \frac{5}{2}, \frac{7}{2} \right) = \left( 2.5, 3.5 \right) \]
-
-
Finding the equations of the medians:
-
Median from A to M_BC (3.5, 2): The slope \( m_{A \to M_{BC}} \) is given by: \[ m = \frac{y_{M_{BC}} - y_A}{x_{M_{BC}} - x_A} = \frac{2 - 5}{3.5 - 4} = \frac{-3}{-0.5} = 6 \] The equation of the line in point-slope form is: \[ y - 5 = 6(x - 4) \]
-
Median from B to M_AC (5, 3.5): The slope \( m_{B \to M_{AC}} \) is: \[ m = \frac{y_{M_{AC}} - y_B}{x_{M_{AC}} - x_B} = \frac{3.5 - 2}{5 - 1} = \frac{1.5}{4} = \frac{3}{8} \] The equation of the line is: \[ y - 2 = \frac{3}{8}(x - 1) \]
-
Median from C to M_AB (2.5, 3.5): The slope \( m_{C \to M_{AB}} \) is: \[ m = \frac{y_{M_{AB}} - y_C}{x_{M_{AB}} - x_C} = \frac{3.5 - 2}{2.5 - 6} = \frac{1.5}{-3.5} = -\frac{3}{7} \] The equation of the line is: \[ y - 2 = -\frac{3}{7}(x - 6) \]
-
-
Finding the intersection point of the medians (the centroid): The centroid (intersection of medians) can also be calculated directly using the coordinates of the vertices: \[ G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} \right) = \left( \frac{4 + 1 + 6}{3}, \frac{5 + 2 + 2}{3} \right) = \left( \frac{11}{3}, \frac{9}{3} \right) = \left( \frac{11}{3}, 3 \right) \]
Thus, the ordered pairs of the midpoints of the sides (which are the points where the medians intersect their respective sides) are:
- Midpoint of BC: (3.5, 2)
- Midpoint of AC: (5, 3.5)
- Midpoint of AB: (2.5, 3.5)
Answers:
- \( M_{BC} = (3.5, 2) \)
- \( M_{AC} = (5, 3.5) \)
- \( M_{AB} = (2.5, 3.5) \)
1 answer