1. **Stating the problem:** Given points A(-2,1), B(2,3), and C(-1,-2), find: a) The equation of the line BC. b) The area of triangle ABC. 2. **Formula and rules:** a) The equation of a line through two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$ b) The area of a triangle with vertices $A(x_1,y_1)$, $B(x_2,y_2)$, and $C(x_3,y_3)$ can be found using the determinant formula: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ 3. **Finding the equation of line BC:** Points B(2,3) and C(-1,-2) Slope $m = \frac{y_C - y_B}{x_C - x_B} = \frac{-2 - 3}{-1 - 2} = \frac{-5}{-3} = \frac{5}{3}$ Using point-slope form with point B: $$y - 3 = \frac{5}{3}(x - 2)$$ Simplify: $$y - 3 = \frac{5}{3}x - \frac{10}{3}$$ $$y = \frac{5}{3}x - \frac{10}{3} + 3 = \frac{5}{3}x - \frac{10}{3} + \frac{9}{3} = \frac{5}{3}x - \frac{1}{3}$$ Equation of line BC is: $$y = \frac{5}{3}x - \frac{1}{3}$$ 4. **Finding the area of triangle ABC:** Using the determinant formula: $$\text{Area} = \frac{1}{2} |(-2)(3 - (-2)) + 2((-2) - 1) + (-1)(1 - 3)|$$ Calculate inside: $$= \frac{1}{2} |(-2)(5) + 2(-3) + (-1)(-2)| = \frac{1}{2} |-10 - 6 + 2| = \frac{1}{2} |-14| = 7$$ **Final answers:** a) The equation of line BC is $y = \frac{5}{3}x - \frac{1}{3}$. b) The area of triangle ABC is 7 square units.