A triangle has vertices \( A(2,-2), B(6,4) \) and \( C(10,1) \). Determine the equation of the sides of the triangle. A parallelogram has vertices \( A(2,1), B(5,4), C(9,4) \) and \( D(6,1) \). Determine the equation of line: \( \begin{array}{llll}\text { a) } A B & \text { b) } B C & \text { c) } C D & \text { d) } A D\end{array} \)
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To find the equations of the sides of the triangle and the parallelogram, we can use the slope-intercept form of the line. For the triangle with vertices \( A(2,-2), B(6,4), C(10,1) \): 1. **Side AB**: The slope \( m \) between A and B can be calculated: \( m_{AB} = \frac{4 - (-2)}{6 - 2} = \frac{6}{4} = 1.5 \). Using the point-slope form, the equation \( y - y_1 = m(x - x_1) \) leads to: \( y + 2 = 1.5(x - 2) \) ⇒ \( y = 1.5x - 5 \). 2. **Side BC**: The slope between B and C: \( m_{BC} = \frac{1 - 4}{10 - 6} = \frac{-3}{4} \). Thus, the equation becomes: \( y - 4 = -\frac{3}{4}(x - 6) \) ⇒ \( y = -\frac{3}{4}x + \frac{33}{4} \). 3. **Side CA**: The slope between C and A: \( m_{CA} = \frac{-2 - 1}{2 - 10} = \frac{-3}{-8} = \frac{3}{8} \). Thus, the equation becomes: \( y - 1 = \frac{3}{8}(x - 10) \) ⇒ \( y = \frac{3}{8}x - \frac{11}{4} \). Now, for the parallelogram with vertices \( A(2,1), B(5,4), C(9,4) \), and \( D(6,1) \): 1. **Side AB**: The slope between A and B: \( m_{AB} = \frac{4 - 1}{5 - 2} = \frac{3}{3} = 1 \). The equation is: \( y - 1 = 1(x - 2) \) ⇒ \( y = x - 1 \). 2. **Side BC**: The slope is \( 0 \) since points B and C have the same y-coordinate: The equation is: \( y = 4 \). 3. **Side CD**: The slope is the same as AB but moving down: Thus, the equation becomes: \( y - 1 = -1(x - 6) \) ⇒ \( y = -x + 7 \). 4. **Side AD**: The slope between A and D: \( m_{AD} = \frac{1 - 1}{2 - 6} = 0 \). The equation is: \( y = 1 \). And there you have the equations for the sides of the triangle and the parallelogram! Happy graphing!
