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GSE Geometry Unit 1 Transformations Review #1 Answer Key

Work through the GSE Geometry Unit 1 Review #1 with angle equations, transversal relationships, and exact coordinate rules for eight transformations.

Question

GSE Geometry Unit 1: Transformations Review #1Solve Problems 1–9 using linear pairs, vertical angles, triangle sums, angle addition, exterior angles, isosceles triangles, complements, and supplements.For Problems 10–13, name each angle pair and state whether the angles are congruent or supplementary. For Problem 14, use m ∥ n and m∠8 = 112° to find ∠1–∠7. Solve the angle equations in Problems 15–16.For Problems 17–24, graph the image of each preimage. Vertices are listed in tracing order; reconnect the final vertex to the first except in Problem 23, which is a line segment.17. (2,9), (0,-1), (6,-7); translate by (x,y) → (x-8,y-3)18. (-10,10), (-8,3), (4,5); reflect across the x-axis19. (-4,7), (2,9), (0,5), (6,-4), (-7,-5), (-1,0); reflect across x=-220. (-9,3), (-5,-2), (-9,-7), (2,-2); reflect across the y-axis21. (-3,-6), (1,-3), (9,-9); rotate 180° about the origin22. (-7,8), (-7,5), (-2,5), (1,8); rotate 90° clockwise about the origin23. Segment endpoints (-8,-2), (-4,9); translate by (x,y) → (x+9,y-8), then rotate 90° counterclockwise about the origin24. (-5,6), (-2,7), (-1,-9); translate by (x,y) → (x+4,y-2), rotate 180° about the origin, then reflect across y=x.

Problems 1–9: angle equations

  1. Linear pair: (10x+1)+(9x−11)=180(10x+1)+(9x-11)=180, so x=10x=10.
  2. Vertical angles: 4x+24=7x+34x+24=7x+3, so x=7x=7.
  3. Triangle sum: 89+57+(5x−6)=18089+57+(5x-6)=180, so x=8x=8.
  4. Angle addition: x+(7x+2)=82x+(7x+2)=82, so x=10x=10.
  5. Exterior-angle theorem: 2x+4=60+x2x+4=60+x, so x=56x=56.
  6. The base angles are both 52∘52^\circ. Thus 14x+6=7614x+6=76, so x=5x=5.
  7. Complementary angles: (12x+4)+(9x+2)=90(12x+4)+(9x+2)=90, so x=4x=4; the angles are 52∘52^\circ and 38∘38^\circ.
  8. Let the smaller supplementary angle be ss. Then s+(s+38)=180s+(s+38)=180, giving 71∘71^\circ and 109∘109^\circ.
  9. Let one angle be aa and its supplement 180−a180-a. From a=2(180−a)−123a=2(180-a)-123, the angles are 79∘79^\circ and 101∘101^\circ.

Problems 10–16: parallel lines and transversals

  1. ∠1\angle1 and ∠5\angle5: corresponding; congruent.
  2. ∠4\angle4 and ∠6\angle6: same-side (consecutive) interior; supplementary.
  3. ∠2\angle2 and ∠8\angle8: alternate exterior; congruent.
  4. ∠4\angle4 and ∠5\angle5: alternate interior; congruent.
  5. If m∥nm\parallel n and m∠8=112∘m\angle8=112^\circ, then m∠1=112∘m\angle1=112^\circ, m∠2=68∘m\angle2=68^\circ, m∠3=68∘m\angle3=68^\circ, m∠4=112∘m\angle4=112^\circ, m∠5=112∘m\angle5=112^\circ, m∠6=68∘m\angle6=68^\circ, and m∠7=68∘m\angle7=68^\circ.
  6. Vertical angles: 3x−50=2x+53x-50=2x+5, so x=55x=55.
  7. Same-side interior angles: (6x+7)+(3x+38)=180(6x+7)+(3x+38)=180, so x=15x=15.

Problems 17–24: transformations

Connect each list of vertices in the order shown. The last vertex reconnects to the first for every closed figure; Problem 23 is a line segment.

ProblemPreimage verticesRuleImage vertices
17(2,9)→(0,−1)→(6,−7)(2,9)\to(0,-1)\to(6,-7)(x,y)↦(x−8,y−3)(x,y)\mapsto(x-8,y-3)(−6,6)→(−8,−4)→(−2,−10)(-6,6)\to(-8,-4)\to(-2,-10)
18(−10,10)→(−8,3)→(4,5)(-10,10)\to(-8,3)\to(4,5)(x,y)↦(x,−y)(x,y)\mapsto(x,-y)(−10,−10)→(−8,−3)→(4,−5)(-10,-10)\to(-8,-3)\to(4,-5)
19(−4,7)→(2,9)→(0,5)→(6,−4)→(−7,−5)→(−1,0)(-4,7)\to(2,9)\to(0,5)\to(6,-4)\to(-7,-5)\to(-1,0)(x,y)↦(−4−x,y)(x,y)\mapsto(-4-x,y)(0,7)→(−6,9)→(−4,5)→(−10,−4)→(3,−5)→(−3,0)(0,7)\to(-6,9)\to(-4,5)\to(-10,-4)\to(3,-5)\to(-3,0)
20(−9,3)→(−5,−2)→(−9,−7)→(2,−2)(-9,3)\to(-5,-2)\to(-9,-7)\to(2,-2)(x,y)↦(−x,y)(x,y)\mapsto(-x,y)(9,3)→(5,−2)→(9,−7)→(−2,−2)(9,3)\to(5,-2)\to(9,-7)\to(-2,-2)
21(−3,−6)→(1,−3)→(9,−9)(-3,-6)\to(1,-3)\to(9,-9)(x,y)↦(−x,−y)(x,y)\mapsto(-x,-y)(3,6)→(−1,3)→(−9,9)(3,6)\to(-1,3)\to(-9,9)
22(−7,8)→(−7,5)→(−2,5)→(1,8)(-7,8)\to(-7,5)\to(-2,5)\to(1,8)(x,y)↦(y,−x)(x,y)\mapsto(y,-x)(8,7)→(5,7)→(5,2)→(8,−1)(8,7)\to(5,7)\to(5,2)\to(8,-1)

Problem 23 composition

The preimage segment has endpoints (−8,−2)(-8,-2) and (−4,9)(-4,9).

  1. Translate by ⟨9,−8⟩\langle9,-8\rangle: (1,−10)(1,-10) and (5,1)(5,1).
  2. Rotate 90∘90^\circ counterclockwise: (10,1)(10,1) and (−1,5)(-1,5).

The combined rule is (x,y)↦(8−y,x+9)(x,y)\mapsto(8-y,x+9).

Problem 24 composition

The preimage triangle has vertices (−5,6)(-5,6), (−2,7)(-2,7), and (−1,−9)(-1,-9).

  1. Translate by ⟨4,−2⟩\langle4,-2\rangle: (−1,4)(-1,4), (2,5)(2,5), and (3,−11)(3,-11).
  2. Rotate 180∘180^\circ: (1,−4)(1,-4), (−2,−5)(-2,-5), and (−3,11)(-3,11).
  3. Reflect across y=xy=x: (−4,1)(-4,1), (−5,−2)(-5,-2), and (11,−3)(11,-3).

The combined rule is (x,y)↦(2−y,−x−4)(x,y)\mapsto(2-y,-x-4).

Completed coordinate graphs for Problems 17–24

17 · Translation (-8, -3)

  • Preimage
  • Final image
17 · Translation (-8, -3)A triangle with preimage vertices A (2, 9), B (0, -1), and C (6, -7), and final image vertices A′ (-6, 6), B′ (-8, -4), and C′ (-2, -10).-12-10-8-6-4-224681012-12-10-8-6-4-224681012xyABCA′B′C′

A triangle with preimage vertices A (2, 9), B (0, -1), and C (6, -7), and final image vertices A′ (-6, 6), B′ (-8, -4), and C′ (-2, -10).

18 · Reflection across the x-axis

  • Preimage
  • Final image
18 · Reflection across the x-axisA triangle with preimage vertices A (-10, 10), B (-8, 3), and C (4, 5), and final image vertices A′ (-10, -10), B′ (-8, -3), and C′ (4, -5).-12-10-8-6-4-224681012-12-10-8-6-4-224681012xyABCA′B′C′

A triangle with preimage vertices A (-10, 10), B (-8, 3), and C (4, 5), and final image vertices A′ (-10, -10), B′ (-8, -3), and C′ (4, -5).

19 · Reflection across x = -2

  • Preimage
  • Final image
19 · Reflection across x = -2A six-vertex preimage A (-4, 7), B (2, 9), C (0, 5), D (6, -4), E (-7, -5), F (-1, 0), and its final reflection A′ (0, 7), B′ (-6, 9), C′ (-4, 5), D′ (-10, -4), E′ (3, -5), F′ (-3, 0).-12-10-8-6-4-224681012-12-10-8-6-4-224681012xyABCDEFA′B′C′D′E′F′

A six-vertex preimage A (-4, 7), B (2, 9), C (0, 5), D (6, -4), E (-7, -5), F (-1, 0), and its final reflection A′ (0, 7), B′ (-6, 9), C′ (-4, 5), D′ (-10, -4), E′ (3, -5), F′ (-3, 0).

20 · Reflection across the y-axis

  • Preimage
  • Final image
20 · Reflection across the y-axisA four-vertex preimage A (-9, 3), B (-5, -2), C (-9, -7), D (2, -2), and final image A′ (9, 3), B′ (5, -2), C′ (9, -7), D′ (-2, -2).-12-10-8-6-4-224681012-12-10-8-6-4-224681012xyABCDA′B′C′D′

A four-vertex preimage A (-9, 3), B (-5, -2), C (-9, -7), D (2, -2), and final image A′ (9, 3), B′ (5, -2), C′ (9, -7), D′ (-2, -2).

21 · Rotation 180° about the origin

  • Preimage
  • Final image
21 · Rotation 180° about the originA triangle with preimage vertices A (-3, -6), B (1, -3), C (9, -9), and final image A′ (3, 6), B′ (-1, 3), C′ (-9, 9).-12-10-8-6-4-224681012-12-10-8-6-4-224681012xyABCA′B′C′

A triangle with preimage vertices A (-3, -6), B (1, -3), C (9, -9), and final image A′ (3, 6), B′ (-1, 3), C′ (-9, 9).

22 · Rotation 90° clockwise

  • Preimage
  • Final image
22 · Rotation 90° clockwiseA four-vertex preimage A (-7, 8), B (-7, 5), C (-2, 5), D (1, 8), and final image A′ (8, 7), B′ (5, 7), C′ (5, 2), D′ (8, -1).-12-10-8-6-4-224681012-12-10-8-6-4-224681012xyABCDA′B′C′D′

A four-vertex preimage A (-7, 8), B (-7, 5), C (-2, 5), D (1, 8), and final image A′ (8, 7), B′ (5, 7), C′ (5, 2), D′ (8, -1).

23 · Translate, then rotate 90° counterclockwise

  • Preimage
  • First intermediate image
  • Final image
23 · Translate, then rotate 90° counterclockwiseA segment moves from preimage endpoints A (-8, -2), B (-4, 9), through first intermediate endpoints A₁ (1, -10), B₁ (5, 1), to final endpoints A′ (10, 1), B′ (-1, 5).-12-10-8-6-4-224681012-12-10-8-6-4-224681012xyABA₁B₁A′B′

A segment moves from preimage endpoints A (-8, -2), B (-4, 9), through first intermediate endpoints A₁ (1, -10), B₁ (5, 1), to final endpoints A′ (10, 1), B′ (-1, 5).

24 · Translate, rotate 180°, then reflect

  • Preimage
  • First intermediate image
  • Second intermediate image
  • Final image
24 · Translate, rotate 180°, then reflectA triangle moves from preimage A (-5, 6), B (-2, 7), C (-1, -9), through translated A₁ (-1, 4), B₁ (2, 5), C₁ (3, -11), then rotated A₂ (1, -4), B₂ (-2, -5), C₂ (-3, 11), to final A′ (-4, 1), B′ (-5, -2), C′ (11, -3).-12-10-8-6-4-224681012-12-10-8-6-4-224681012xyABCA₁B₁C₁A₂B₂C₂A′B′C′

A triangle moves from preimage A (-5, 6), B (-2, 7), C (-1, -9), through translated A₁ (-1, 4), B₁ (2, 5), C₁ (3, -11), then rotated A₂ (1, -4), B₂ (-2, -5), C₂ (-3, 11), to final A′ (-4, 1), B′ (-5, -2), C′ (11, -3).

Evidence boundary

Problems 1–16 were reconstructed from the blank review and teacher-key copies. For Problems 17–24, every preimage vertex and transformation step was checked against a clear Cobb County School District answer key whose shared figures are confirmed by the matching vertices in Problems 17 and 21; every image coordinate was then recalculated from the stated rule.

Sources

These references support the concepts and methods used in the explanation above.

GSE Geometry Unit 1 Transformations Review Answer Key | Verla