Math Problem Statement

Rotate a trialg with coordinates A(7,15), B(5,8), C(10,10) about (5,8) and scale it by (2,2) about (10,10)

Solution

To solve this problem, we'll approach it step-by-step:

  1. Rotation about point B(5,8):
    To rotate the triangle by a given angle about a point, we first translate the point to the origin, rotate, and then translate back. However, you haven't provided a specific rotation angle.

    If you meant a 90° counterclockwise rotation (a common choice), here's how we'd do it:

    The formula for a 90° counterclockwise rotation about a point (x0,y0)(x_0, y_0) is: (x,y)=(y0(yy0),x0+(xx0))(x', y') = (y_0 - (y - y_0), x_0 + (x - x_0)) So for each vertex, we will:

    1. Translate by subtracting point B(5,8)B(5,8).
    2. Apply the 90° rotation.
    3. Translate back to point B(5,8)B(5,8).
  2. Scaling about point C(10,10):
    To scale by a factor of (2,2)(2,2) about point C(10,10)C(10,10), we use the formula: (x,y)=(Cx+sx(xCx),Cy+sy(yCy))(x', y') = (C_x + s_x \cdot (x - C_x), C_y + s_y \cdot (y - C_y)) Where sx=2s_x = 2 and sy=2s_y = 2 (scaling factors). We apply this scaling to each rotated vertex.


Step 1: Rotate Triangle about Point B(5,8)B(5,8)

Let's first calculate the rotated positions (assuming a 90° counterclockwise rotation) of the vertices of the triangle about B(5,8)B(5,8):

  1. Vertex A(7, 15): A=(5+(158),8(75))=(5+7,82)=(12,6)A' = (5 + (15 - 8), 8 - (7 - 5)) = (5 + 7, 8 - 2) = (12, 6)

  2. Vertex B(5, 8):
    This is the center of rotation, so it stays the same: B=(5,8)B' = (5, 8)

  3. Vertex C(10, 10): C=(5+(108),8(105))=(5+2,85)=(7,3)C' = (5 + (10 - 8), 8 - (10 - 5)) = (5 + 2, 8 - 5) = (7, 3)

So, the rotated triangle has vertices at A(12,6)A'(12,6), B(5,8)B'(5,8), and C(7,3)C'(7,3).


Step 2: Scale about Point C(10,10)C(10,10)

Now, we will scale the rotated triangle about C(10,10)C(10,10) by a factor of (2,2)(2,2):

  1. Vertex A(12,6)A'(12, 6): A=(10+2(1210),10+2(610))=(10+4,10+(8))=(14,2)A'' = (10 + 2 \cdot (12 - 10), 10 + 2 \cdot (6 - 10)) = (10 + 4, 10 + (-8)) = (14, 2)

  2. Vertex B(5,8)B'(5, 8): B=(10+2(510),10+2(810))=(10+(10),10+(4))=(0,6)B'' = (10 + 2 \cdot (5 - 10), 10 + 2 \cdot (8 - 10)) = (10 + (-10), 10 + (-4)) = (0, 6)

  3. Vertex C(7,3)C'(7, 3): C=(10+2(710),10+2(310))=(10+(6),10+(14))=(4,4)C'' = (10 + 2 \cdot (7 - 10), 10 + 2 \cdot (3 - 10)) = (10 + (-6), 10 + (-14)) = (4, -4)


Final Answer:

The vertices of the triangle after a 90° counterclockwise rotation about B(5,8)B(5,8) and scaling by (2,2)(2,2) about C(10,10)C(10,10) are:

  • A(14,2)A''(14, 2)
  • B(0,6)B''(0, 6)
  • C(4,4)C''(4, -4)

Would you like more details about how to perform these transformations, or do you have any specific questions?

Follow-up Questions:

  1. How do rotation transformations work in 2D?
  2. What happens if the scaling factor is not uniform (different for xx and yy)?
  3. How can you apply a rotation for an arbitrary angle?
  4. What is the effect of scaling by a factor less than 1?
  5. How do translation and rotation interact geometrically?

Tip:

When performing transformations, always work step by step (translate, rotate, then translate back) for clarity and accuracy.

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Math Problem Analysis

Mathematical Concepts

Geometry
Coordinate Transformations
Rotation
Scaling

Formulas

(x', y') = (y_0 - (y - y_0), x_0 + (x - x_0))
(x', y') = (C_x + s_x * (x - C_x), C_y + s_y * (y - C_y))

Theorems

Rotation Theorem
Scaling Theorem

Suitable Grade Level

Grades 9-11