Math Problem Statement

The coordinates of the points A and B are (-3,1) and (3,9) respectively. A is rotated anticlockwise about the point P(1, 2) through 180° to A’. (a) Write down the coordinates of A’. (b) Find the distance between A and B. (c) Yoda claims that it is possible to obtain the point C(5, 6) by rotating the point A about P with some degrees. Do you agree? Explain your answer.

Solution

Let's break down and solve the given problem step by step. Here's the problem:

  1. Given Information:
    • Coordinates of points A(3,1)A(-3, 1) and B(3,9)B(3, 9).
    • Point AA is rotated anticlockwise about point P(1,2)P(1, 2) through 180180^\circ.

Solution for Each Subpart:

(a) Write down the coordinates of AA'.

To rotate a point (x,y)(x, y) about another point (h,k)(h, k) by 180180^\circ:

  • The new coordinates are: A(x,y)=(2hx,2ky).A'(x', y') = (2h - x, 2k - y).

Here, A(3,1)A(-3, 1) is rotated about P(1,2)P(1, 2):

  • x=2(1)(3)=2+3=5x' = 2(1) - (-3) = 2 + 3 = 5,
  • y=2(2)1=41=3y' = 2(2) - 1 = 4 - 1 = 3.

So, the coordinates of AA' are: A(5,3).A'(5, 3).


(b) Find the distance between AA and BB.

The distance dd between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by: d=(x2x1)2+(y2y1)2.d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Substitute A(3,1)A(-3, 1) and B(3,9)B(3, 9): d=(3(3))2+(91)2=(3+3)2+(8)2=62+82=36+64=100=10.d = \sqrt{(3 - (-3))^2 + (9 - 1)^2} = \sqrt{(3 + 3)^2 + (8)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10.

So, the distance between AA and BB is: 10units.10 \, \text{units}.


(c) Yoda claims it is possible to obtain the point C(5,6)C(5, 6) by rotating the point A(3,1)A(-3, 1) about P(1,2)P(1, 2) with some degrees. Do you agree?

We test whether C(5,6)C(5, 6) can be obtained by a rotation about P(1,2)P(1, 2):

  1. First, calculate the vector from PP to AA: PA=(31,12)=(4,1).PA = (-3 - 1, 1 - 2) = (-4, -1).

  2. Next, calculate the vector from PP to CC: PC=(51,62)=(4,4).PC = (5 - 1, 6 - 2) = (4, 4).

  3. For C(5,6)C(5, 6) to be a rotation of A(3,1)A(-3, 1) about P(1,2)P(1, 2), the length of PAPA and PCPC must be equal, and the angle between them must correspond to a standard rotation.

    • Length of PAPA: PA=(4)2+(1)2=16+1=17.|PA| = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}.

    • Length of PCPC: PC=(4)2+(4)2=16+16=32.|PC| = \sqrt{(4)^2 + (4)^2} = \sqrt{16 + 16} = \sqrt{32}.

Since PAPC|PA| \neq |PC|, it is not possible to obtain C(5,6)C(5, 6) by rotating A(3,1)A(-3, 1) about P(1,2)P(1, 2).

Conclusion: Yoda’s claim is incorrect.


Summary of Answers:

  • (a) A(5,3)A'(5, 3).
  • (b) Distance between AA and BB is 10units10 \, \text{units}.
  • (c) It is not possible to obtain C(5,6)C(5, 6) by rotating A(3,1)A(-3, 1) about P(1,2)P(1, 2).

Would you like detailed steps for any part? Let me know!


Further Questions for Practice:

  1. How would the coordinates change if the rotation was 9090^\circ instead of 180180^\circ?
  2. How can you find the angle of rotation needed to map AA to CC?
  3. What are the coordinates of BB' if B(3,9)B(3, 9) is rotated 180180^\circ about P(1,2)P(1, 2)?
  4. How would you determine the midpoint of ABAB?
  5. Can you verify if A,B,A', B, and CC form a specific geometric figure?

Tip: Always visualize rotations on a coordinate plane to better understand transformations!

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Distance Formula
Rotation of Points

Formulas

Rotation formula for 180°: (x', y') = (2h - x, 2k - y)
Distance formula: d = √((x2 - x1)^2 + (y2 - y1)^2)

Theorems

Properties of Rotations
Euclidean Distance

Suitable Grade Level

Grades 9-12