Question

NO LINKS!!

Part 3
Graph the polygon with the given vertices and its image after the transformation. Label all vertices in both the image and preimage using the correct notation.

7. A(-4, 3), B(-5, -1), C(0, 2)
Translation:(x, y)--> (x + 2, y + 1)

8. A(-2, -4), B(-2, 4), C(2, 2)
Dilation: k = 1/2

Answer 1

Apply the indicated rule for each vertex of the polygon and then graph both the image and preimage.

Question 7

• A(-4, 3), B(-5, -1), C(0, 2)
• Translation:(x, y) → (x + 2, y + 1)

• A(-4, 3) → A'(- 4 + 2, 3 + 1 ) = A'(-2, 4)
• B(-5, -1) → B'(-5 + 2, - 1 + 1) = B'(-3, 0)
• C(0, 2) → C'(0 + 2, 2 + 1) = C'(2, 3)

Question 8

• A(-2, -4), B(-2, 4), C(2, 2)
• Dilation: k = 1/2

• A(-2, - 4) → A'(-2/2, - 4/2) = A'(-1, - 2)
• B(-2, 4) → B'(-2/2, 4/2) = B'(-1, 2)
• C(2, 2) → C'(2/2, 2/2) = C'(1, 1)

See attached for both.

Answer 2

Question 7:

• A' (-2, 4)
• B' (-3, 0)
• C' (2, 3)

Question 8:

• A' (-1, -2)
• B' (-1, 2)
• C' (1, 2)

Explanation:

Question 7

Given vertices of the pre-image:

• A = (-4, 3)
• B = (-5, -1)
• C = (0, 2)

Given transformation rule:

• 
  `(x, y) \rightarrow (x + 2, y + 1)`

Therefore the vertices of the image are:

• A' = (-4 + 2, 3 + 1) = (-2, 4)
• B' = (-5 + 2, -1 + 1) = (-3, 0)
• C' = (0 + 2, 2 + 1) = (2, 3)

Question 8

Given vertices of the pre-image:

• A = (-2, -4)
• B = (-2, 4)
• C = (2, 2)

If the pre-image is dilated by a scale factor of ¹/₂ then the transformation rule is:

• 
  `(x, y) \rightarrow \left((1)/(2)x, (1)/(2)y\right)`

Therefore the vertices of the image are:

• 
  `\sf A' = \left((1)/(2) \cdot -2, (1)/(2) \cdot -4\right)= \left(-1, -2\right)`
• 
  `\sf B' = \left((1)/(2) \cdot -2, (1)/(2) \cdot 4\right) = \left(-1, 2\right)`
• 
  `\sf C' = \left((1)/(2) \cdot 2, (1)/(2) \cdot 2\right) = \left(1, 1\right)`