Geometry Diff Chapter 3 Test Review Answers 1. Alternate Exterior Angles 2. Consecutive Interior Angles 3. Alternate Interior Angles

4.

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m =−6 −18 − 8

=−70

, therefore the slope is undefined

5.

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m =0 − 64 − 0

=−64

= −32

6.

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m =3 − 36 − −6( )

=012

= 0

7.

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m =1− 48 − 5

=−33

= −1

8.

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m∠9 = 84°, because ∠9 is a consecutive interior angle with

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∠8, and

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∠8 is a vertical angle with

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∠2. 9.

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m∠11 =138°, because ∠11 is a linear pair angle with

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∠12. 10.

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m∠6 = 42° , because ∠6 is an alternate interior angle with

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∠12. 11. Because the two angles marked are alternate exterior angles, then x −8 =120⇒ x =128 13. For the line

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y = 2x −17,

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m = 2 , then the

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⊥ m = − 12 , therefore the equation of perpendicular line is

y−1= − 12 x − −8( )( )⇒ y−1= − 12 x − 4⇒ y = − 12 x −3 14. For the line

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y = 4x −19,

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m = 4 , then the parallel m = 4 , therefore the equation of parallel line is y− 7 = 4 x − 0( )⇒ y− 7 = 4x − 0⇒ y = 4x + 7

15. For the line

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y = − 23 x −11,

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m = − 23 , then the

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⊥ m = 32 , therefore the equation of perpendicular line is

y−3= 32 x − −12( )( )⇒ y−3= 32 x +18⇒ y = 32 x + 21 16. For the line

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y = x −11, use the point B(0, -11) and

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m =1, then the

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⊥ m = −1, giving the equation of perpendicular line as y =mx + b⇒ y = −x −11 . The intersection of the lines

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y = x − 7 and

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y = −x −11 is x − 7 = −x −11⇒ 2x = −4⇒ x = −2 and y = x − 7⇒ y = −2( )− 7⇒ y = −9 , giving the intersection point (-2, -9).

The distance from (0, -11) to (-2, -9) is d = −2− 0( )2 + −9− −11( )( )2 = −2( )2 + 2( )2 = 8 ≈ 2.828 .

Graph for problem #16. Graph for problem #17.

17. For the line

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y = −2x +1, use the point B(0, 1) and

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m = −2 , then the

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⊥ m = 12 , giving equation of

perpendicular line as y =mx + b⇒ y = 12 x +1 .

The intersection of the lines

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y = −2x +16 and

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y = 12 x +1 is −2x +16 = 12 x +1⇒ 2 −2x +16 = 12 x +1( )

⇒−4x +32 = x + 2⇒−5x = −30⇒ x = 6 and y = −2x +16⇒ y = −2 6( )+16⇒ y = 4 , giving the intersection point (6, 4).

The distance from (0, 1) to (6, 4) is

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d = 6 − 0( )2 + 4 −1( )2 = 6( )2 + 3( )2 = 45 ≈ 6.708.

18. D, because the lines containing

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CD and

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VZ will never intersect and are not on the same plane. 19. Because the two angles marked are consecutive interior angles, then

4x +11+8x +1=180⇒12x +12 =180⇒12x =168⇒ x =14 .

20. The slope between the point (-4, 2) and (3, -5) is

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m =−5 − 23 − −4( )

=−77

= −1. Using the point (-4, 2), then the

equation between those points as y− 2 = −1 x − −4( )( )⇒ y− 2 = −x − 4⇒ y = −x − 2 .

Then the

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⊥ m =1 and the perpendicular line through the point (1, 2) is y− 2 =1 x −1( )⇒ y− 2 = x −1⇒ y = x +1 . The intersection of the lines

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y = −x − 2 and

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y = x +1 is −x − 2 = x +1⇒−2x = 3⇒ x = − 32 and

y = x +1⇒ y = − 32 +1⇒ y = − 12 , giving the intersection point − 32 , −12( ) .

The distance from (1, 2) to

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− 32 , − 1

2( ) is

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d = − 32 −1( )2 + − 12 − 2( )2 = − 52( )2 + − 52( )2 = 252 ≈ 3.536.

Graph for problem #20. Graph for problem #21.

21. The slope between the point (6, 5) and (2, 3) is

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m =3 − 52 − 6

=−2−4

=12

. Using the point (6, 5), then the

equation between those points is y− 5= 12 x − 6( )⇒ y− 5= 12 x −3⇒ y = 12 x + 2 .

Then the

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⊥ m = −2 and the perpendicular line through the point (2, 6) is y− 6 = −2 x − 2( )⇒ y− 6 = −2x + 4⇒ y = −2x +10 . The intersection of the lines

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y = 12 x + 2 and

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y = −2x +10 is 12 x + 2 = −2x +10⇒ 2 1

2 x + 2 = −2x +10( )⇒ x + 4 = −4x + 20⇒ 5x =16⇒ x = 165 and

y = −2x +10⇒ y = −2 165( )+10⇒ y = − 325 +

505 =

185 , giving intersection point 165 , 

185( ) .

The distance from (2, 6) to

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165 , 18

5( ) is

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d = 165 − 2( )2 + 18

5 − 6( )2 = 65( )2 + −125( )2 = 180

25 ≈ 2.683.

22.

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j || k by the Corresponding Angles Converse Postulate. 23. No lines are parallel because there is no common transversal. 24.

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p ||q by the Alternate Exterior Angles Converse Theorem.