Graph △ABC with vertices A(1, 2), B(3, 4), and C(2, 0). Perform the following sequence of transformations: Step 1: Translate △ABC using the vector to obtain △A′B′C′ Step 2: Reflect △A′B′C′ in the y-axis to obtain △A″B″C″ Step 3: Rotate △A″B″C″ through 90° anticlockwise about the origin to obtain △A‴B‴C‴ Step 4: Enlarge △A‴B‴C‴ with a scale factor of k = 2 centred at the origin to obtain △A⁴B⁴C⁴ Show your work clearly, including: The coordinate rule and coordinates for each transformation step A labelled graph showing the original triangle and all four images (△ABC, △A′B′C′, △A″B″C″, △A‴B‴C‴, △A⁴B⁴C⁴) A comparison of the perimeter of △ABC and △A⁴B⁴C⁴

The coordinate rule for translation by vector $\begin{pmatrix} a \\ b \end{pmatrix}$ is $(x, y) \to (x+a, y+b)$.
For this step, the rule is $(x, y) \to (x-4, y-3)$.

$$A' = (1-4, 2-3) = (-3, -1)$$ $$B' = (3-4, 4-3) = (-1, 1)$$ $$C' = (2-4, 0-3) = (-2, -3)$$

The coordinate rule for reflection in the y-axis is $(x, y) \to (-x, y)$.

$$A'' = (-(-3), -1) = (3, -1)$$ $$B'' = (-(-1), 1) = (1, 1)$$ $$C'' = (-(-2), -3) = (2, -3)$$

The coordinate rule for a 90° anticlockwise rotation about the origin is $(x, y) \to (-y, x)$.

$$A''' = (-(-1), 3) = (1, 3)$$ $$B''' = (-(1), 1) = (-1, 1)$$ $$C''' = (-(-3), 2) = (3, 2)$$

The coordinate rule for enlargement with scale factor $k$ centred at the origin is $(x, y) \to (kx, ky)$.
For this step, the rule is $(x, y) \to (2x, 2y)$.

$$A^4 = (2 \cdot 1, 2 \cdot 3) = (2, 6)$$ $$B^4 = (2 \cdot (-1), 2 \cdot 1) = (-2, 2)$$ $$C^4 = (2 \cdot 3, 2 \cdot 2) = (6, 4)$$

To graph, plot the following vertices for each triangle on a coordinate plane and connect them:

• △ABC: A(1, 2), B(3, 4), C(2, 0)
• △A′B′C′: A′(-3, -1), B′(-1, 1), C′(-2, -3)
• △A″B″C″: A″(3, -1), B″(1, 1), C″(2, -3)
• △A‴B‴C‴: A‴(1, 3), B‴(-1, 1), C‴(3, 2)
• △A⁴B⁴C⁴: A⁴(2, 6), B⁴(-2, 2), C⁴(6, 4)

The distance formula is $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

$$AB = \sqrt{(3-1)^2 + (4-2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4+4} = \sqrt{8}$$ $$BC = \sqrt{(2-3)^2 + (0-4)^2} = \sqrt{(-1)^2 + (-4)^2} = \sqrt{1+16} = \sqrt{17}$$ $$CA = \sqrt{(1-2)^2 + (2-0)^2} = \sqrt{(-1)^2 + 2^2} = \sqrt{1+4} = \sqrt{5}$$

Perimeter of △ABC $= \sqrt{8} + \sqrt{17} + \sqrt{5}$