Answer :
To determine whether sides [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] of the triangle with vertices [tex]\( A(-2,2) \)[/tex], [tex]\( B(1,5) \)[/tex], and [tex]\( C(6,-1) \)[/tex] are parallel or perpendicular, we need to calculate the slopes of these two segments.
1. Calculate the slope of line segment [tex]\( AB \)[/tex]:
- The formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
- For points [tex]\( A(-2, 2) \)[/tex] and [tex]\( B(1, 5) \)[/tex]:
[tex]\[ \text{slope}_{AB} = \frac{5 - 2}{1 - (-2)} = \frac{3}{3} = 1 \][/tex]
Thus, the slope of [tex]\( AB \)[/tex] is [tex]\( 1 \)[/tex].
2. Calculate the slope of line segment [tex]\( BC \)[/tex]:
- For points [tex]\( B(1, 5) \)[/tex] and [tex]\( C(6, -1) \)[/tex]:
[tex]\[ \text{slope}_{BC} = \frac{-1 - 5}{6 - 1} = \frac{-6}{5} = -1.2 \][/tex]
Thus, the slope of [tex]\( BC \)[/tex] is [tex]\( -1.2 \)[/tex].
3. Determine if the sides are parallel:
- Lines are parallel if and only if their slopes are equal.
- Here, we have:
[tex]\[ \text{slope}_{AB} = 1 \quad \text{and} \quad \text{slope}_{BC} = -1.2 \][/tex]
- Since [tex]\( 1 \neq -1.2 \)[/tex], sides [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] are not parallel.
4. Determine if the sides are perpendicular:
- Lines are perpendicular if and only if the product of their slopes is [tex]\(-1\)[/tex].
- For our slopes:
[tex]\[ \text{slope}_{AB} \times \text{slope}_{BC} = 1 \times (-1.2) = -1.2 \][/tex]
- Since [tex]\( -1.2 \neq -1 \)[/tex], sides [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] are not perpendicular.
Therefore, the sides [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] of the triangle with vertices [tex]\( A(-2, 2) \)[/tex], [tex]\( B(1, 5) \)[/tex], and [tex]\( C(6, -1) \)[/tex] are neither parallel nor perpendicular.
1. Calculate the slope of line segment [tex]\( AB \)[/tex]:
- The formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
- For points [tex]\( A(-2, 2) \)[/tex] and [tex]\( B(1, 5) \)[/tex]:
[tex]\[ \text{slope}_{AB} = \frac{5 - 2}{1 - (-2)} = \frac{3}{3} = 1 \][/tex]
Thus, the slope of [tex]\( AB \)[/tex] is [tex]\( 1 \)[/tex].
2. Calculate the slope of line segment [tex]\( BC \)[/tex]:
- For points [tex]\( B(1, 5) \)[/tex] and [tex]\( C(6, -1) \)[/tex]:
[tex]\[ \text{slope}_{BC} = \frac{-1 - 5}{6 - 1} = \frac{-6}{5} = -1.2 \][/tex]
Thus, the slope of [tex]\( BC \)[/tex] is [tex]\( -1.2 \)[/tex].
3. Determine if the sides are parallel:
- Lines are parallel if and only if their slopes are equal.
- Here, we have:
[tex]\[ \text{slope}_{AB} = 1 \quad \text{and} \quad \text{slope}_{BC} = -1.2 \][/tex]
- Since [tex]\( 1 \neq -1.2 \)[/tex], sides [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] are not parallel.
4. Determine if the sides are perpendicular:
- Lines are perpendicular if and only if the product of their slopes is [tex]\(-1\)[/tex].
- For our slopes:
[tex]\[ \text{slope}_{AB} \times \text{slope}_{BC} = 1 \times (-1.2) = -1.2 \][/tex]
- Since [tex]\( -1.2 \neq -1 \)[/tex], sides [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] are not perpendicular.
Therefore, the sides [tex]\( AB \)[/tex] and [tex]\( BC \)[/tex] of the triangle with vertices [tex]\( A(-2, 2) \)[/tex], [tex]\( B(1, 5) \)[/tex], and [tex]\( C(6, -1) \)[/tex] are neither parallel nor perpendicular.
