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What relections or sequence of reflections transforms figure ABCD into figure EFGH?

E(-5,1)
F(-2,5)
G(-2,8)
H(-4,7)
A(1,-5)
B(5,-2)
C(8,-2)
D(7,-4)

The answer choices are:

Reflect ABCD first over the line x=0, then over the line y=0
Reflect ABCD over the line y=x
Reflect ABCD first over the line x=0 then over the line y=x
Reflect ABCD first over the line y=0 then over the line y=x

3 answers

To determine which sequence of reflections transforms figure ABCD into figure EFGH, let's first analyze the coordinates of the points in both figures.

Coordinates of ABCD:

  • A(1, -5)
  • B(5, -2)
  • C(8, -2)
  • D(7, -4)

Coordinates of EFGH:

  • E(-5, 1)
  • F(-2, 5)
  • G(-2, 8)
  • H(-4, 7)

Now, let's calculate the result of each proposed transformation and see which one results in the points of EFGH.

  1. Reflect over the line \(x = 0\) (the y-axis):

    • A(1, -5) → A'(-1, -5)
    • B(5, -2) → B'(-5, -2)
    • C(8, -2) → C'(-8, -2)
    • D(7, -4) → D'(-7, -4)

    Resulting points after first reflection:

    • A'(-1, -5)
    • B'(-5, -2)
    • C'(-8, -2)
    • D'(-7, -4)

  2. Next, reflect the resulting points over the line \(y = 0\) (the x-axis):

    • A'(-1, -5) → A''(-1, 5)
    • B'(-5, -2) → B''(-5, 2)
    • C'(-8, -2) → C''(-8, 2)
    • D'(-7, -4) → D''(-7, 4)

    Resulting points after second reflection:

    • A''(-1, 5)
    • B''(-5, 2)
    • C''(-8, 2)
    • D''(-7, 4)

This result does not match EFGH.

  1. Reflect over the line \(y = x\):

    • A(1, -5) → A'(−5, 1)
    • B(5, -2) → B'(−2, 5)
    • C(8, -2) → C'(−2, 8)
    • D(7, -4) → D'(−4, 7)

    Resulting points after reflection over y = x:

    • A'(-5, 1)
    • B'(-2, 5)
    • C'(-2, 8)
    • D'(-4, 7)

This result matches EFGH perfectly: E(-5,1), F(-2,5), G(-2,8), H(-4,7).

  1. Reflection over the line \(y = 0\) then over the line \(y = x\):

    • Reflecting A, B, C, D over the y-axis results in A'(-1, -5), B'(-5, -2), C'(-8, -2), D'(-7, -4) (as already calculated).

    • Then applying reflection over y=x gives:

    • A'(-1, -5) → A''(-5, -1)

    • B'(-5, -2) → B''(-2, -5)

    • C'(-8, -2) → C''(-2, -8)

    • D'(-7, -4) → D''(-4, -7)

    This does not match EFGH.

  2. Reflect over y=0 then y=x: We already calculated that this does not yield EFGH.

The only successful transformation that matches is reflecting ABCD over the line \(y = x\).

Thus, the answer is:

Reflect ABCD over the line \(y=x\).

Are you sure that's right

Let's carefully re-evaluate the transformation step by step to confirm the accurate reflection that transforms figure \(ABCD\) into figure \(EFGH\).

Given Points

  • Figure \(ABCD\):

    • \(A(1, -5)\)
    • \(B(5, -2)\)
    • \(C(8, -2)\)
    • \(D(7, -4)\)

  • Figure \(EFGH\):

    • \(E(-5, 1)\)
    • \(F(-2, 5)\)
    • \(G(-2, 8)\)
    • \(H(-4, 7)\)

Analysis

  1. Initial Reflection Calculations: Let's explore the transformations one step at a time.

Option 1: Reflect over the line \(x = 0\) then \(y = 0\)

  • Reflecting over the line \(x=0\) (y-axis):

    • \(A(1, -5) \rightarrow A'(-1, -5)\)
    • \(B(5, -2) \rightarrow B'(-5, -2)\)
    • \(C(8, -2) \rightarrow C'(-8, -2)\)
    • \(D(7, -4) \rightarrow D'(-7, -4)\)

  • Reflecting those over the line \(y=0\) (x-axis):

    • \(A'(-1, -5) \rightarrow A''(-1, 5)\)
    • \(B'(-5, -2) \rightarrow B''(-5, 2)\)
    • \(C'(-8, -2) \rightarrow C''(-8, 2)\)
    • \(D'(-7, -4) \rightarrow D''(-7, 4)\)

Resulting points: \(A''(-1, 5)\), \(B''(-5, 2)\), \(C''(-8, 2)\), \(D''(-7, 4)\) — This does not match \(EFGH\).

Option 2: Reflect over the line \(y = x\)

  • Reflecting over the line \(y = x\):
    • \(A(1, -5) \rightarrow A'(-5, 1)\)
    • \(B(5, -2) \rightarrow B'(-2, 5)\)
    • \(C(8, -2) \rightarrow C'(-2, 8)\)
    • \(D(7, -4) \rightarrow D'(-4, 7)\)

Resulting points: \(A'(-5, 1)\), \(B'(-2, 5)\), \(C'(-2, 8)\), \(D'(-4, 7)\) — This matches \(EFGH\).

Other Options

  1. Option 3: Reflect over \(x=0\) then \(y=x\)

  • Reflect over \(x=0\):

    • \(A(1, -5) \rightarrow A'(-1, -5)\)
    • \(B(5, -2) \rightarrow B'(-5, -2)\)
    • \(C(8, -2) \rightarrow C'(-8, -2)\)
    • \(D(7, -4) \rightarrow D'(-7, -4)\)

  • Reflect those over \(y=x\):

    • \(A'(-1, -5) \rightarrow A''(-5, -1)\)
    • \(B'(-5, -2) \rightarrow B''(-2, -5)\)
    • \(C'(-8, -2) \rightarrow C''(-2, -8)\)
    • \(D'(-7, -4) \rightarrow D''(-4, -7)\)

Resulting points: \(A''(-5, -1)\), \(B''(-2, -5)\), \(C''(-2, -8)\), \(D''(-4, -7)\) — This does not match \(EFGH\).

  1. Option 4: Reflect over \(y=0\) then \(y=x\)

  • Over \(y=0\) results in:

    • \(A(-1, 5)\), similar to above.

  • Then reflect over the line \(y=x\), similar to above, also does not yield \(EFGH\).

Conclusion

The correct transformation is indeed: Reflect \(ABCD\) over the line \(y=x\), as this directly transforms the coordinates of \(ABCD\) into those of \(EFGH\). Thank you for your patience!