Directions: Find the centroid of each triangle. \( \begin{array}{l}\text { 1. } \mathrm{A}(3,-2), \mathrm{B}(0,6) \text { and } \mathrm{C}(-3,-2) \\ \text { 3. } \mathrm{U}(-1,-5), \mathrm{V}(-4,0) \text { and } \mathrm{W}(-2,-3) \\ \text { 5. } \mathrm{P}(1,-4), \mathrm{Q}(0,5) \text { and } \mathrm{R}(-9,10) \\ \text { 7. } \mathrm{P}(6,6), \mathrm{Q}(1,4) \text { and } \mathrm{R}(5,2) \\ \text { 4. } \mathrm{X}(-1,0), \mathrm{Y}(3,5) \text { and } \mathrm{Z}(4,-2) \\ \text { 6. } \mathrm{A}(2,3), \mathrm{B}(1,7) \text { and } \mathrm{C}(4,5) \\ \text { 8. } \mathrm{S}(7,8), \mathrm{T}(-4,3) \text { and } \mathrm{U}(0,10)\end{array} \)

To find the centroid (also known as the geometric center) of a triangle formed by points \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\), you can use the formula: \[ \text{Centroid} \, (G) = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right) \] Now, let's calculate the centroids for the given triangles: 1. For triangle \(ABC\) with vertices \(A(3,-2)\), \(B(0,6)\), and \(C(-3,-2)\): \[ G_1 = \left( \frac{3 + 0 - 3}{3}, \frac{-2 + 6 - 2}{3} \right) = \left(0, \frac{2}{3}\right) \] 3. For triangle \(UVW\) with vertices \(U(-1,-5)\), \(V(-4,0)\), and \(W(-2,-3)\): \[ G_3 = \left( \frac{-1 - 4 - 2}{3}, \frac{-5 + 0 - 3}{3} \right) = \left(-2.33, -2.67\right) \] 5. For triangle \(PQR\) with vertices \(P(1,-4)\), \(Q(0,5)\), and \(R(-9,10)\): \[ G_5 = \left( \frac{1 + 0 - 9}{3}, \frac{-4 + 5 + 10}{3} \right) = \left(-2.67, 3.67\right) \] 7. For triangle \(PQR\) with vertices \(P(6,6)\), \(Q(1,4)\), and \(R(5,2)\): \[ G_7 = \left( \frac{6 + 1 + 5}{3}, \frac{6 + 4 + 2}{3} \right) = \left(4, 4\right) \] 4. For triangle \(XYZ\) with vertices \(X(-1,0)\), \(Y(3,5)\), and \(Z(4,-2)\): \[ G_4 = \left( \frac{-1 + 3 + 4}{3}, \frac{0 + 5 - 2}{3} \right) = \left(2, 1\right) \] 6. For triangle \(ABC\) with vertices \(A(2,3)\), \(B(1,7)\), and \(C(4,5)\): \[ G_6 = \left( \frac{2 + 1 + 4}{3}, \frac{3 + 7 + 5}{3} \right) = \left(2.33, 5\right) \] 8. For triangle \(STU\) with vertices \(S(7,8)\), \(T(-4,3)\), and \(U(0,10)\): \[ G_8 = \left( \frac{7 - 4 + 0}{3}, \frac{8 + 3 + 10}{3} \right) = \left(1, 7\right) \] So, the centroids for the triangles are: - Triangle ABC: \( (0, \frac{2}{3}) \) - Triangle UVW: \( (-2.33, -2.67) \) - Triangle PQR: \( (-2.67, 3.67) \) - Triangle PQR: \( (4, 4) \) - Triangle XYZ: \( (2, 1) \) - Triangle ABC: \( (2.33, 5) \) - Triangle STU: \( (1, 7) \)