ChristopherLothmanGeometryDistanceFor
Christopher Lothman
Geometry
Distance Formula:
Translation A’ (10, 1) B’ (2, 1) C’ (2, 7)
Rule: (x, y) > (x 8, y + 4) A2C2 = sqrt(((2 (10))^2) + ((7 1)^2)) = 10
AC = sqrt(((6 (2))^2) + ((3 (3))^2)) = 10
A2C2 = AC
A2B2 = sqrt(((2 (10))^2) + ((1 1)^2)) = 8
AB = sqrt(((6 (2))^2) + ((3 (3))^2)) = 8
A2B2 = AB
B2C2 = sqrt(((2 (2))^2) + ((7 1)^2)) = 6
BC = sqrt(((6 6)^2) + ((3 (3))^2)) = 6
B2C2 = BC
Triangle A2B2C2 = Triangle ABC by construction. Reflection A’ (4, 1) B’ (4, 6) C’ (4, 6)
Line of reflection is x=y
Angle A2 = Angle C by construction.
Angle C2 = Angle A by construction.
AC = sqrt(((4 (4))^2) + ((5 (1))^2)) = 10
A2C2 = sqrt(((6 (2))^2) + ((3 ( 3))^2)) = 10
AC = A2C2
Triangle A2C2B2 = Triangle ACB by construction. Rotation A’ (2, 3) B’ (6, 3) C’ (6, 3)
A2C2 = sqrt(((6 2)^2) + ((3 3)^2)) = 10
AC = sqrt(((6 (2))^2) + ((3 (3))^2)) = 10
A2C2 = AC
A2B2 = sqrt(((6 2)^2) + ((3 3)^2)) = 8
AB = sqrt(((6 (2))^2) + ((3 (3))^2)) = 8
A2B2 = AB
Angle A2 = Angle A by construction.
Triangle B2A2C2 = Triangle BAC by construction.
Provide answers to the following questions along with your transformations:
1. Describe the translation you performed on the original triangle. Use details and coordinates to explain how the figure was transformed. Be sure to use complete sentences in your answer.
I moved the triangle across the x and y axis into the 1st quadrant making the new coordinates A’ (10, 1), B’ (2, 1), and C’ (2, 7).
2. How many degrees did you rotate your triangle? In which direction
(clockwise, counterclockwise) did it move? Be sure to use complete sentences in your answer.
I rotated my triangle 180 degrees counter clockwise in to the 1st quadrant.
3. What line of reflection did you choose for your transformation? How are you sure that each point was reflected across
Geometry
Distance Formula:
Translation A’ (10, 1) B’ (2, 1) C’ (2, 7)
Rule: (x, y) > (x 8, y + 4) A2C2 = sqrt(((2 (10))^2) + ((7 1)^2)) = 10
AC = sqrt(((6 (2))^2) + ((3 (3))^2)) = 10
A2C2 = AC
A2B2 = sqrt(((2 (10))^2) + ((1 1)^2)) = 8
AB = sqrt(((6 (2))^2) + ((3 (3))^2)) = 8
A2B2 = AB
B2C2 = sqrt(((2 (2))^2) + ((7 1)^2)) = 6
BC = sqrt(((6 6)^2) + ((3 (3))^2)) = 6
B2C2 = BC
Triangle A2B2C2 = Triangle ABC by construction. Reflection A’ (4, 1) B’ (4, 6) C’ (4, 6)
Line of reflection is x=y
Angle A2 = Angle C by construction.
Angle C2 = Angle A by construction.
AC = sqrt(((4 (4))^2) + ((5 (1))^2)) = 10
A2C2 = sqrt(((6 (2))^2) + ((3 ( 3))^2)) = 10
AC = A2C2
Triangle A2C2B2 = Triangle ACB by construction. Rotation A’ (2, 3) B’ (6, 3) C’ (6, 3)
A2C2 = sqrt(((6 2)^2) + ((3 3)^2)) = 10
AC = sqrt(((6 (2))^2) + ((3 (3))^2)) = 10
A2C2 = AC
A2B2 = sqrt(((6 2)^2) + ((3 3)^2)) = 8
AB = sqrt(((6 (2))^2) + ((3 (3))^2)) = 8
A2B2 = AB
Angle A2 = Angle A by construction.
Triangle B2A2C2 = Triangle BAC by construction.
Provide answers to the following questions along with your transformations:
1. Describe the translation you performed on the original triangle. Use details and coordinates to explain how the figure was transformed. Be sure to use complete sentences in your answer.
I moved the triangle across the x and y axis into the 1st quadrant making the new coordinates A’ (10, 1), B’ (2, 1), and C’ (2, 7).
2. How many degrees did you rotate your triangle? In which direction
(clockwise, counterclockwise) did it move? Be sure to use complete sentences in your answer.
I rotated my triangle 180 degrees counter clockwise in to the 1st quadrant.
3. What line of reflection did you choose for your transformation? How are you sure that each point was reflected across