Question

6 sarah draws figure efgh in the coordinate plane. she rotates figure efgh 90° clockwise around vertex e to form image efgh. draw the image efgh on the coordinate plane.

Explanation:

Step1: Identify Coordinates

First, find coordinates of \( E, F, G, H \). From graph:
\( E(1, -1) \), \( F(3, -2) \), \( G(3, -4) \), \( H(1, -4) \).

Step2: Rotation Rule (90° Clockwise)

For a point \( (x, y) \) rotated 90° clockwise around \( (a, b) \), the formula is:
\( (x', y') = (a + (y - b), b - (x - a)) \).
Since we rotate around \( E(1, -1) \), \( a = 1 \), \( b = -1 \).

Step3: Rotate \( F \)

For \( F(3, -2) \):
\( x' = 1 + (-2 - (-1)) = 1 - 1 = 0 \)
\( y' = -1 - (3 - 1) = -1 - 2 = -3 \)
So \( F'(0, -3) \).

Step4: Rotate \( G \)

For \( G(3, -4) \):
\( x' = 1 + (-4 - (-1)) = 1 - 3 = -2 \)
\( y' = -1 - (3 - 1) = -1 - 2 = -3 \)
So \( G'(-2, -3) \).

Step5: Rotate \( H \)

For \( H(1, -4) \):
\( x' = 1 + (-4 - (-1)) = 1 - 3 = -2 \)
\( y' = -1 - (1 - 1) = -1 - 0 = -1 \)
So \( H'(-2, -1) \).

Step6: Plot Points

Plot \( E(1, -1) \), \( F'(0, -3) \), \( G'(-2, -3) \), \( H'(-2, -1) \), then connect to form \( EF'G'H' \).

(Note: Since this is a drawing task, the key is calculating rotated coordinates. The final figure connects these points: \( E \) to \( F' \) to \( G' \) to \( H' \) to \( E \).)

Answer:

To draw \( EF'G'H' \), plot the rotated points:

• \( E(1, -1) \) (unchanged, center of rotation),
• \( F'(0, -3) \),
• \( G'(-2, -3) \),
• \( H'(-2, -1) \),

then connect the points in order.