91Ó°ÊÓ

To find the area of a triangle when you have its vertices, you can use a special formula. For a triangle with points \(A(x_1, y_1)\), \(B(x_2, y_2)\), \(C(x_3, y_3)\), the formula for the area is: \[ \text{Area} = \frac{1}{2} \, |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|. \] Now let's break down the problem:
Given points: \(A(-1,-2)\), \(B(5,1)\), \(C(7,-3)\), substitute these into the formula: \[ \text{Area} = \frac{1}{2} | -1(1 - (-3)) + 5((-3) - (-2)) + 7((-2) - 1)| = \frac{1}{2} | -1(4) + 5(-1) + 7(-3)| = \frac{1}{2} | -4 - 5 - 21| = \frac{1}{2} | -30| = 15. \]
So, the area of our triangle is 15.

The centroid of a triangle is the point where its three medians intersect. A median is a line from a vertex to the midpoint of the opposite side. The coordinates of the centroid \(G(x,y)\) of a triangle with vertices \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\) are calculated as follows: \[ G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right).\] For our triangle with points \(A(-1,-2)\), \(B(5,1)\), and \(C(7,-3)\), we find the centroid by substituting the coordinates:
\[ G \left( \frac{-1+5+7}{3}, \frac{-2+1-3}{3} \right) = \left( \frac{11}{3}, \frac{-4}{3} \right). \] Therefore, the centroid of the triangle is \( \left(\frac{11}{3}, \frac{-4}{3} \right) \).

The height of a triangle can be found if you know the area and the length of the side corresponding to that height. The relationship used here is: \[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}.\] Let's find the heights corresponding to each vertex.

**1. Height from Vertex A**
Using the side BC as the base:
Calculate the length of BC:
\[ BC = \sqrt{(7-5)^2 + (-3-1)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{20} = 2\sqrt{5}.\] Now use the area formula to find the height from A:
\[ 15 = \frac{1}{2} \times 2\sqrt{5} \times \text{height}\] \[ \text{height}_A = \frac{15}{\sqrt{5}} = 3\sqrt{5}.\]

**2. Height from Vertex B**
Using the side AC as the base:
Calculate the length of AC:
\[ AC = \sqrt{(7-(-1))^2 + (-3-(-2))^2} = \sqrt{8^2 + (-1)^2} = \sqrt{65}.\] Now use the area formula to find the height from B:
\[15 = \frac{1}{2} \times \sqrt{65} \times \text{height} \] \[\text{height}_B = \frac{30}{\sqrt{65}} = \frac{30\sqrt{65}}{65} = \frac{6\sqrt{65}}{13}.\]

**3. Height from Vertex C**
Using the side AB as the base:
Calculate the length of AB:
\[ AB = \sqrt{(5 - (-1))^2 + (1-(-2))^2} = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}. \] Now use the area formula to find the height from C:
\[ 15 = \frac{1}{2} \times 3\sqrt{5} \times \text{height} \] \[ \text{height}_C = \frac{15}{1.5\sqrt{5}} = \frac{15}{1.5} \times \frac{1}{\sqrt{5}} = \frac{10}{\sqrt{5}} = 2\sqrt{5}. \]