Question

1. rhombus defg with vertices d(-7, 8), e(-2, 7), f(-1, 2), and g(-6, 3): 90° counterclockwise about (2, 1) d’( , ) e’( , ) f’( , ) g’( , )

Explanation:

To rotate a point \((x, y)\) \(90^\circ\) counterclockwise about a center \((h, k)\), we use the following steps:
1. Translate the point so that the center is at the origin: \((x - h, y - k)\)
2. Rotate the translated point \(90^\circ\) counterclockwise: \((-(y - k), x - h)\)
3. Translate back by adding the center coordinates: \((h - (y - k), k + (x - h))\) or simplified \((h - y + k, k + x - h)\)

Step 1: Rotate \(D(-7, 8)\) about \((2, 1)\)

- Translate: \((-7 - 2, 8 - 1) = (-9, 7)\)
- Rotate \(90^\circ\) CCW: \((-7, -9)\)
- Translate back: \((2 - 7, 1 - 9) = (-5, -8)\)? Wait, no, wait the formula is \((h - (y - k), k + (x - h))\). So \(h = 2\), \(k = 1\), \(x = -7\), \(y = 8\)
\(h - (y - k) = 2 - (8 - 1) = 2 - 7 = -5\)
\(k + (x - h) = 1 + (-7 - 2) = 1 - 9 = -8\)? Wait, that seems off. Wait, the correct rotation of a point \((a, b)\) \(90^\circ\) counterclockwise about the origin is \((-b, a)\). So after translating, the point is \((a, b) = (x - h, y - k)\), then rotated is \((-b, a)\), then translated back is \((-b + h, a + k)\). Let's redo:

For \(D(-7, 8)\), center \((2, 1)\):
- Translate: \(a = -7 - 2 = -9\), \(b = 8 - 1 = 7\)
- Rotate \(90^\circ\) CCW: \((-b, a) = (-7, -9)\)
- Translate back: \((-7 + 2, -9 + 1) = (-5, -8)\). Wait, but let's check with another method. Let's use the rotation formula for a point \((x,y)\) around \((h,k)\): the new coordinates \((x', y')\) are given by:
\(x' = h - (y - k)\)
\(y' = k + (x - h)\)

So for \(D(-7,8)\):
\(x' = 2 - (8 - 1) = 2 - 7 = -5\)
\(y' = 1 + (-7 - 2) = 1 - 9 = -8\). So \(D'(-5, -8)\)

Step 2: Rotate \(E(-2, 7)\) about \((2, 1)\)

- Translate: \(a = -2 - 2 = -4\), \(b = 7 - 1 = 6\)
- Rotate \(90^\circ\) CCW: \((-6, -4)\)
- Translate back: \((-6 + 2, -4 + 1) = (-4, -3)\)
Using the formula:
\(x' = 2 - (7 - 1) = 2 - 6 = -4\)
\(y' = 1 + (-2 - 2) = 1 - 4 = -3\). So \(E'(-4, -3)\)

Step 3: Rotate \(F(-1, 2)\) about \((2, 1)\)

- Translate: \(a = -1 - 2 = -3\), \(b = 2 - 1 = 1\)
- Rotate \(90^\circ\) CCW: \((-1, -3)\)
- Translate back: \((-1 + 2, -3 + 1) = (1, -2)\)
Using the formula:
\(x' = 2 - (2 - 1) = 2 - 1 = 1\)
\(y' = 1 + (-1 - 2) = 1 - 3 = -2\). So \(F'(1, -2)\)

Step 4: Rotate \(G(-6, 3)\) about \((2, 1)\)

- Translate: \(a = -6 - 2 = -8\), \(b = 3 - 1 = 2\)
- Rotate \(90^\circ\) CCW: \((-2, -8)\)
- Translate back: \((-2 + 2, -8 + 1) = (0, -7)\)
Using the formula:
\(x' = 2 - (3 - 1) = 2 - 2 = 0\)
\(y' = 1 + (-6 - 2) = 1 - 8 = -7\). So \(G'(0, -7)\)

Wait, but let's verify with a different approach. Let's plot the points or use vector rotation.

Alternatively, the rotation of a point \((x, y)\) 90 degrees counterclockwise about \((h, k)\) can be calculated as:

\(x' = h - (y - k)\)

\(y' = k + (x - h)\)

Let's recalculate each:

For \(D(-7,8)\):

\(x' = 2 - (8 - 1) = 2 - 7 = -5\)

\(y' = 1 + (-7 - 2) = 1 - 9 = -8\) → \(D'(-5, -8)\)

For \(E(-2,7)\):

\(x' = 2 - (7 - 1) = 2 - 6 = -4\)

\(y' = 1 + (-2 - 2) = 1 - 4 = -3\) → \(E'(-4, -3)\)

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

\(x' = 2 - (2 - 1) = 2 - 1 = 1\)

\(y' = 1 + (-1 - 2) = 1 - 3 = -2\) → \(F'(1, -2)\)

For \(G(-6,3)\):

\(x' = 2 - (3 - 1) = 2 - 2 = 0\)

\(y' = 1 + (-6 - 2) = 1 - 8 = -7\) → \(G'(0, -7)\)

Let's check if these make sense. Let's take vector from center to D: \((-7 - 2, 8 - 1) = (-9,7)\). Rotating this vector 90 degrees counterclockwise: \((-7, -9)\) (since (a,b) → (-b,a)). Then adding the center: \((-7 + 2, -9 + 1) = (-5, -8)\), which matches. Good.

For E: vector from center: (-4,6). Rotate 90 CCW: (-6, -4). Add center: (-6 + 2, -4 + 1) = (-4, -3). Correct.

For F: vector from center: (-3,1). Rotate 90 CCW: (-1, -3). Add center: (-1 + 2, -3 + 1) = (1, -2). Correct.

For G: vector from center: (-8,2). Rotate 90 CCW: (-2, -8). Add center: (-2 + 2, -8 + 1) = (0, -7). Correct.

Answer:

\(D'(-5, -8)\)
\(E'(-4, -3)\)
\(F'(1, -2)\)
\(G'(0, -7)\)