Question

1. rotation 90° clockwise about the origin

Explanation:

To solve the problem of rotating triangle \( FGH \) \( 90^\circ \) clockwise about the origin, we first need to determine the coordinates of the vertices \( F \), \( G \), and \( H \). Let's assume the grid has each square with side length 1.

Step 1: Identify Coordinates of Vertices

• Let's find the coordinates of \( F \), \( G \), and \( H \):
• \( F \): Let's say \( F \) is at \( (-5, -2) \) (assuming from the grid, we need to check the exact position. Wait, looking at the grid, let's re-examine:
• \( H \) is at \( (0, -3) \) (since it's on the y-axis, 3 units down from origin)
• \( G \) is at \( (-2, -1) \) (2 units left on x, 1 unit down on y)
• \( F \) is at \( (-5, -2) \) (5 units left on x, 2 units down on y)

Wait, maybe better to get exact coordinates. Let's assume:

• \( F \): Let's count the grid. From origin (0,0), moving left 5, down 2: \( F(-5, -2) \)
• \( G \): Left 2, down 1: \( G(-2, -1) \)
• \( H \): At (0, -3) (since it's on the y-axis, 3 units below origin)

Step 2: Apply \( 90^\circ \) Clockwise Rotation Formula

The formula for rotating a point \( (x, y) \) \( 90^\circ \) clockwise about the origin is:
\[
(x, y)
ightarrow (y, -x)
\]

Let's apply this to each vertex:

For \( F(-5, -2) \):

Using the formula \( (x, y)
ightarrow (y, -x) \):
\( x = -5 \), \( y = -2 \)
New coordinates: \( (-2, -(-5)) = (-2, 5) \)

For \( G(-2, -1) \):

\( x = -2 \), \( y = -1 \)
New coordinates: \( (-1, -(-2)) = (-1, 2) \)

For \( H(0, -3) \):

\( x = 0 \), \( y = -3 \)
New coordinates: \( (-3, -0) = (-3, 0) \)

Step 3: Plot the New Points

Now, we plot the new points \( F'(-2, 5) \), \( G'(-1, 2) \), and \( H'(-3, 0) \) on the grid and connect them to form the rotated triangle.

Final Answer

The coordinates of the rotated triangle \( F'G'H' \) after \( 90^\circ \) clockwise rotation about the origin are:

• \( F'(-2, 5) \)
• \( G'(-1, 2) \)
• \( H'(-3, 0) \)

(Note: If the initial coordinates were different, the rotated coordinates would adjust accordingly. The key is applying the \( 90^\circ \) clockwise rotation formula \( (x, y)
ightarrow (y, -x) \) to each vertex.)

Answer:

To solve the problem of rotating triangle \( FGH \) \( 90^\circ \) clockwise about the origin, we first need to determine the coordinates of the vertices \( F \), \( G \), and \( H \). Let's assume the grid has each square with side length 1.

Step 1: Identify Coordinates of Vertices

• Let's find the coordinates of \( F \), \( G \), and \( H \):
• \( F \): Let's say \( F \) is at \( (-5, -2) \) (assuming from the grid, we need to check the exact position. Wait, looking at the grid, let's re-examine:
• \( H \) is at \( (0, -3) \) (since it's on the y-axis, 3 units down from origin)
• \( G \) is at \( (-2, -1) \) (2 units left on x, 1 unit down on y)
• \( F \) is at \( (-5, -2) \) (5 units left on x, 2 units down on y)

Wait, maybe better to get exact coordinates. Let's assume:

• \( F \): Let's count the grid. From origin (0,0), moving left 5, down 2: \( F(-5, -2) \)
• \( G \): Left 2, down 1: \( G(-2, -1) \)
• \( H \): At (0, -3) (since it's on the y-axis, 3 units below origin)

Step 2: Apply \( 90^\circ \) Clockwise Rotation Formula

The formula for rotating a point \( (x, y) \) \( 90^\circ \) clockwise about the origin is:
\[
(x, y)
ightarrow (y, -x)
\]

Let's apply this to each vertex:

For \( F(-5, -2) \):

Using the formula \( (x, y)
ightarrow (y, -x) \):
\( x = -5 \), \( y = -2 \)
New coordinates: \( (-2, -(-5)) = (-2, 5) \)

For \( G(-2, -1) \):

\( x = -2 \), \( y = -1 \)
New coordinates: \( (-1, -(-2)) = (-1, 2) \)

For \( H(0, -3) \):

\( x = 0 \), \( y = -3 \)
New coordinates: \( (-3, -0) = (-3, 0) \)

Step 3: Plot the New Points

Now, we plot the new points \( F'(-2, 5) \), \( G'(-1, 2) \), and \( H'(-3, 0) \) on the grid and connect them to form the rotated triangle.

Final Answer

The coordinates of the rotated triangle \( F'G'H' \) after \( 90^\circ \) clockwise rotation about the origin are:

• \( F'(-2, 5) \)
• \( G'(-1, 2) \)
• \( H'(-3, 0) \)

(Note: If the initial coordinates were different, the rotated coordinates would adjust accordingly. The key is applying the \( 90^\circ \) clockwise rotation formula \( (x, y)
ightarrow (y, -x) \) to each vertex.)