Question

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graph the image of \\(\delta fgh\\) after a rotation \\(90^\circ\\) counterclockwise around the origin and a reflection over the \\(y\\)-axis.

Explanation:

Identify original coordinates

Find the coordinates of the vertices of \(\Delta FGH\) from the graph:

• \(F = (2, -2)\)
• \(G = (3, -6)\)
• \(H = (1, -6)\)

Apply rotation transformation

Using Coordinate Rotation of \(90^\circ\) counterclockwise around the origin:

$$(x, y) ightarrow (-y, x)$$

• \(F(2, -2)

ightarrow F'(-(-2), 2) = (2, 2)\)

• \(G(3, -6)

ightarrow G'(-(-6), 3) = (6, 3)\)

• \(H(1, -6)

ightarrow H'(-(-6), 1) = (6, 1)\)

Apply reflection transformation

Using Reflection over the \(y\)-axis:

$$(x, y) ightarrow (-x, y)$$

• \(F'(2, 2)

ightarrow F''(-2, 2)\)

• \(G'(6, 3)

ightarrow G''(-6, 3)\)

• \(H'(6, 1)

ightarrow H''(-6, 1)\)

Verify final coordinates

The final vertices of the transformed triangle \(\Delta F''G''H''\) are:

• \(F'' = (-2, 2)\)
• \(G'' = (-6, 3)\)
• \(H'' = (-6, 1)\)

This matches the plotted solid triangle labeled \(\Delta F'G'H'\) in the image.

Answer:

The coordinates of the final image after a \(90^\circ\) counterclockwise rotation around the origin followed by a reflection over the \(y\)-axis are:

• \(F'' = (-2, 2)\)
• \(G'' = (-6, 3)\)
• \(H'' = (-6, 1)\)