Question

The coordinates of the vertices of triangle ABC are A(-2,4), B(-7,-1), and C(-3,-3). Prove that triangle ABC is isosceles

State the coordinates of triangle A' B' C', the image of triangle ABC, after a translation 5 units to the right and 5 units down

Answer 1

By calculating side lengths using the distance formula, we proved triangle ABC is isosceles as sides AB and CA are equal. The coordinates of the vertices of triangle A'B'C' after the translation are A'(3, -1), B'(-2, -6), and C'(2, -8).

Step-by-step explanation:

To prove that triangle ABC is isosceles, we need to show that two sides of the triangle have equal length. We can use the distance formula d = √((x2-x1)² + (y2-y1)²) to calculate the lengths of sides AB, BC, and CA.

Length of AB = √((-7 - (-2))² + (-1 - 4)²) = √(25 + 25) = √(50)

Length of BC = √((-3 - (-7))² + (-3 - (-1))²) = √(16 + 4) = √(20)

Length of CA = √((-2 - (-3))² + (4 - (-3))²) = √(1 + 49) = √(50)

Sides AB and CA both have length √(50), proving that triangle ABC is isosceles.

The image of triangle ABC after a translation 5 units to the right and 5 units down (A' B' C') will have its vertices at the following coordinates:

A' = (-2 + 5, 4 - 5) = (3, -1)

B' = (-7 + 5, -1 - 5) = (-2, -6)

C' = (-3 + 5, -3 - 5) = (2, -8)

Answer 2

The image of triangle ABC after the translation 5 units to the right and 5 units down is A'B'C' with vertices at (3, -1), (-2, -6), and (2, -8).

Step-by-step explanation:

To prove that triangle ABC is isosceles, we need to show that at least two of its sides have the same length. We can use the distance formula to calculate the length of each side:

AB = sqrt[(−7−(−2))^2 + (−1−4)^2] = sqrt[25 + 25] = 5sqrt(2)

AC = sqrt[(−3−(−2))^2 + (−3−4)^2] = sqrt[1 + 49] = sqrt(50)

BC = sqrt[(−7−(−3))^2 + (−1−(−3))^2] = sqrt[16 + 4] = 2sqrt(10)

Since AB and AC have different lengths, triangle ABC cannot be equilateral. However, if we compare AB and BC, we see that they have the same length, 5sqrt(2). Therefore, triangle ABC is isosceles.

To find the coordinates of A'B'C' after the translation 5 units to the right and 5 units down, we simply add 5 to the x-coordinates and subtract 5 from the y-coordinates of each vertex. Thus:

A' = (-2+5, 4-5) = (3, -1)

B' = (-7+5, -1-5) = (-2, -6)

C' = (-3+5, -3-5) = (2, -8)

Therefore, the image of triangle ABC after the translation 5 units to the right and 5 units down is A'B'C' with vertices at (3, -1), (-2, -6), and (2, -8).