Respuesta :
We'll use rise over run to find the slopes of two of the sides of the triangle:
[tex] \frac{y_{2} - y_{1}}{x_{2} - x_{1}}[/tex]
Because C is the rightmost point, we'll have it represent (x1,y1), and have A and B represent (x2,y2):
A to C
(-2,3) to (2,-1)
[tex]\frac{-1 - 3}{2 + 2} = \frac{-4}{4} = -1[/tex]
The slope of the line from A to C is -1.
B to C
(-3,-6) to (2,-1)
[tex]\frac{-1 + 6}{2+3} = \frac{5}{5} = 1[/tex]
The slope of the line from B to C is 1.
These lines intersect at (2,-1), and they have negative reciprocal slopes of each other, which creates a perpendicular angle. This means that triangle ABC is a right triangle, and angle ACB is the right angle.
[tex] \frac{y_{2} - y_{1}}{x_{2} - x_{1}}[/tex]
Because C is the rightmost point, we'll have it represent (x1,y1), and have A and B represent (x2,y2):
A to C
(-2,3) to (2,-1)
[tex]\frac{-1 - 3}{2 + 2} = \frac{-4}{4} = -1[/tex]
The slope of the line from A to C is -1.
B to C
(-3,-6) to (2,-1)
[tex]\frac{-1 + 6}{2+3} = \frac{5}{5} = 1[/tex]
The slope of the line from B to C is 1.
These lines intersect at (2,-1), and they have negative reciprocal slopes of each other, which creates a perpendicular angle. This means that triangle ABC is a right triangle, and angle ACB is the right angle.
Answer:
B. A’(3, –2), B’(3, –5), C’(2, –4)
Step-by-step explanation:
Got it right on Edge :)