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Find the coordinates of the centroid of each triangle given the three vertices. \( \begin{array}{l}\text { 27) } \mathrm{C}(-7,-6), \mathrm{B}(-8,0), \mathrm{A}(0,0)\end{array} \) \( \begin{array}{ll}\text { 29) } \mathrm{T}(0,0), \mathrm{U}(1,3), \mathrm{V}(-2,0) & \text { 30) } \mathrm{Y}(3,4), \mathrm{X}(-5,0), \mathrm{W}(0,0)\end{array} \) \( \begin{array}{ll}\text { 31) } \mathrm{E}(6,-3), \mathrm{D}(-5,-3), \mathrm{C}(6,-1)\end{array} \)

Ask by Wells Gibson. in the United States
Mar 12,2025

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Answer

**Centroids of the Triangles:** 1. **Triangle 27:** - Centroid: \((-5, -2)\) 2. **Triangle 29:** - Centroid: \(\left(-\frac{1}{3}, 1\right)\) 3. **Triangle 30:** - Centroid: \(\left(-\frac{2}{3}, \frac{4}{3}\right)\) 4. **Triangle 31:** - Centroid: \(\left(\frac{7}{3}, -\frac{7}{3}\right)\)

Solution

**Triangle 27: Vertices \( C(-7,-6) \), \( B(-8,0) \), \( A(0,0) \)** 1. Write the formula for the centroid: \[ \left( \frac{x_1+x_2+x_3}{3},\, \frac{y_1+y_2+y_3}{3} \right) \] 2. Substitute the coordinates: \[ \text{x-coordinate: } \frac{-7+(-8)+0}{3} = \frac{-15}{3} = -5 \] \[ \text{y-coordinate: } \frac{-6+0+0}{3} = \frac{-6}{3} = -2 \] 3. The centroid is: \[ (-5,\,-2) \] --- **Triangle 29: Vertices \( T(0,0) \), \( U(1,3) \), \( V(-2,0) \)** 1. Use the same centroid formula: \[ \left( \frac{x_1+x_2+x_3}{3},\, \frac{y_1+y_2+y_3}{3} \right) \] 2. Compute the coordinates: \[ \text{x-coordinate: } \frac{0+1+(-2)}{3} = \frac{-1}{3} = -\frac{1}{3} \] \[ \text{y-coordinate: } \frac{0+3+0}{3} = \frac{3}{3} = 1 \] 3. The centroid is: \[ \left(-\frac{1}{3},\, 1\right) \] --- **Triangle 30: Vertices \( Y(3,4) \), \( X(-5,0) \), \( W(0,0) \)** 1. Using the centroid formula: \[ \left( \frac{x_1+x_2+x_3}{3},\, \frac{y_1+y_2+y_3}{3} \right) \] 2. Compute the coordinates: \[ \text{x-coordinate: } \frac{3+(-5)+0}{3} = \frac{-2}{3} = -\frac{2}{3} \] \[ \text{y-coordinate: } \frac{4+0+0}{3} = \frac{4}{3} \] 3. The centroid is: \[ \left(-\frac{2}{3},\, \frac{4}{3}\right) \] --- **Triangle 31: Vertices \( E(6,-3) \), \( D(-5,-3) \), \( C(6,-1) \)** 1. Apply the centroid formula: \[ \left( \frac{x_1+x_2+x_3}{3},\, \frac{y_1+y_2+y_3}{3} \right) \] 2. Compute the coordinates: \[ \text{x-coordinate: } \frac{6+(-5)+6}{3} = \frac{7}{3} = \frac{7}{3} \] \[ \text{y-coordinate: } \frac{-3+(-3)+(-1)}{3} = \frac{-7}{3} = -\frac{7}{3} \] 3. The centroid is: \[ \left(\frac{7}{3},\, -\frac{7}{3}\right) \]

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Bonus Knowledge

To find the coordinates of the centroid of a triangle given its vertices, you can use the formula: \[ \text{Centroid } (G) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \] For each set of vertices, let's calculate the centroid: 1) For triangle C(-7,-6), B(-8,0), A(0,0): \[ G = \left( \frac{-7 + (-8) + 0}{3}, \frac{-6 + 0 + 0}{3} \right) = \left( \frac{-15}{3}, \frac{-6}{3} \right) = (-5,-2) \] 2) For triangle T(0,0), U(1,3), V(-2,0): \[ G = \left( \frac{0 + 1 + (-2)}{3}, \frac{0 + 3 + 0}{3} \right) = \left( \frac{-1}{3}, 1 \right) \] 3) For triangle Y(3,4), X(-5,0), W(0,0): \[ G = \left( \frac{3 + (-5) + 0}{3}, \frac{4 + 0 + 0}{3} \right) = \left( \frac{-2}{3}, \frac{4}{3} \right) \] 4) For triangle E(6,-3), D(-5,-3), C(6,-1): \[ G = \left( \frac{6 + (-5) + 6}{3}, \frac{-3 + (-3) + (-1)}{3} \right) = \left( \frac{7}{3}, \frac{-7}{3} \right) \] So, the centroids of the triangles are: 1) Triangle CAB: Centroid at (-5,-2) 2) Triangle TUV: Centroid at \((- \frac{1}{3}, 1)\) 3) Triangle YXW: Centroid at \((- \frac{2}{3}, \frac{4}{3})\) 4) Triangle EDC: Centroid at \((\frac{7}{3}, -\frac{7}{3})\)

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