Examkin

Question

Difficulty: HardTransformations in the Coordinate Plane

In the standard (x,y)(x, y) coordinate plane, line segment ABAB has endpoints A(−2,4)A(-2, 4) and B(4,2)B(4, 2). The segment is rotated 90∘90^\circ counterclockwise about the origin, then reflected across the line y=−xy = -x, and finally translated 33 units to the right and 22 units down. What are the coordinates of the midpoint of the final image of the segment?

  1. A
    (−4,5)(-4, 5)
  2. B
    (−3,6)(-3, 6)
  3. C
    (4,−5)(4, -5)
  4. (2,1)(2, 1)Answer
  5. E
    (0,−1)(0, -1)

Answer

(2,1)(2, 1)
The correct midpoint is (2,1)(2, 1). First, the midpoint of the original segment ABAB is calculated using the midpoint formula, yielding (1,3)(1, 3). Rotating this point 90∘90^\circ counterclockwise about the origin maps (x,y)(x, y) to (−y,x)(-y, x), which gives (−3,1)(-3, 1). Reflecting (−3,1)(-3, 1) across the line y=−xy = -x maps (x,y)(x, y) to (−y,−x)(-y, -x), resulting in (−1,3)(-1, 3). Finally, translating this point 33 units to the right and 22 units down yields (−1+3,3−2)=(2,1)(-1 + 3, 3 - 2) = (2, 1).

Step-by-Step Solution

1
Find the midpoint of the original line segment ABAB.
M(1,3)M(1, 3)
The midpoint formula is M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right). Substituting the coordinates of A(−2,4)A(-2, 4) and B(4,2)B(4, 2) yields (−2+42,4+22)=(1,3)\left(\frac{-2 + 4}{2}, \frac{4 + 2}{2}\right) = (1, 3).
2
Apply a 90∘90^\circ counterclockwise rotation about the origin to the midpoint.
M′(−3,1)M'(-3, 1)
A 90∘90^\circ counterclockwise rotation about the origin maps (x,y)→(−y,x)(x, y) \rightarrow (-y, x). Applying this rule to M(1,3)M(1, 3) gives (−3,1)(-3, 1).
3
Reflect the rotated point across the line y=−xy = -x.
M′′(−1,3)M''(-1, 3)
A reflection across the line y=−xy = -x maps (x,y)→(−y,−x)(x, y) \rightarrow (-y, -x). Applying this rule to M′(−3,1)M'(-3, 1) gives (−1,3)(-1, 3).
4
Translate the reflected point 33 units right and 22 units down.
M′′′(2,1)M'''(2, 1)
A translation of 33 units right and 22 units down maps (x,y)→(x+3,y−2)(x, y) \rightarrow (x + 3, y - 2). Translating M′′(−1,3)M''(-1, 3) yields (−1+3,3−2)=(2,1)(-1 + 3, 3 - 2) = (2, 1).

Key Concept

Composite transformations of geometric figures in the coordinate plane

Alternative Method

Instead of transforming the midpoint, you can transform the individual endpoints A(−2,4)A(-2, 4) and B(4,2)B(4, 2) through all three steps. Point A(−2,4)A(-2, 4) becomes A′(−4,−2)→A′′(2,4)→A′′′(5,2)A'(-4, -2) \rightarrow A''(2, 4) \rightarrow A'''(5, 2). Point B(4,2)B(4, 2) becomes B′(−2,4)→B′′(−4,2)→B′′′(−1,0)B'(-2, 4) \rightarrow B''(-4, 2) \rightarrow B'''(-1, 0). The midpoint of A′′′A''' and B′′′B''' is then calculated to be (5+(−1)2,2+02)=(2,1)\left(\frac{5 + (-1)}{2}, \frac{2 + 0}{2}\right) = (2, 1).
Estimated Time:1m 30s
Rate this question

Topics

Transformations in the Coordinate Plane

More questions from this topic