12 multiple-choice questions, progressively harder.
Triangle ABCABCABC has A(−3,2)A(-3,2)A(−3,2), B(1,4)B(1,4)B(1,4), and C(2,1)C(2,1)C(2,1). Reflect it across the xxx-axis. Which list gives A′,B′,C′A',B',C'A′,B′,C′ in that order?
Solution
Correct answer: B
Keep each horizontal coordinate and reverse each height.
A′(−3,−2)A'(-3,-2)A′(−3,−2)
B′(1,−4)B'(1,-4)B′(1,−4)
C′(2,−1)C'(2,-1)C′(2,−1)
Join corresponding vertices in the original order.
A reflection in a vertical line sends E(−8,6)E(-8,6)E(−8,6) to E′(4,6)E'(4,6)E′(4,6). What is the mirror?
Correct answer: D
The mirror's horizontal coordinate is halfway between the two horizontal coordinates.
h=−8+42=−2h=\frac{-8+4}{2}=-2h=2−8+4=−2
The connector is horizontal, so x=−2x=-2x=−2 is perpendicular to it and passes through its midpoint.
A reflection in a horizontal line sends F(7,−9)F(7,-9)F(7,−9) to F′(7,1)F'(7,1)F′(7,1). What is the mirror?
Correct answer: A
Average the two heights to locate the midpoint of the vertical connector.
k=−9+12=−4k=\frac{-9+1}{2}=-4k=2−9+1=−4
The horizontal mirror is y=−4y=-4y=−4.
Reflection across y=−xy=-xy=−x sends PPP to P′(3,−8)P'(3,-8)P′(3,−8). Where did PPP start?
Correct answer: C
The image rule is (x,y)↦(−y,−x)(x,y)\mapsto(-y,-x)(x,y)↦(−y,−x). For the image's first coordinate, −y=3-y=3−y=3, so y=−3y=-3y=−3. For its second, −x=−8-x=-8−x=−8, so x=8x=8x=8.
(8,−3)↦(3,−8)(8,-3)\mapsto(3,-8)(8,−3)↦(3,−8)
Point W(t,7)W(t,7)W(t,7) is fixed by reflection across x=−6x=-6x=−6. What is ttt?
A fixed point must be on the mirror. Every point of the line x=−6x=-6x=−6 has first coordinate -6.
t=−6t=-6t=−6
Reflection maps ABABAB to A′B′A'B'A′B′. Given AB=7u−4AB=7u-4AB=7u−4 and A′B′=38A'B'=38A′B′=38, find uuu.
Corresponding distances are equal.
7u−4=387u-4=387u−4=38
7u=427u=427u=42
u=6u=6u=6
Substitution gives length 7(6)−4=387(6)-4=387(6)−4=38, as required.
A reflection sends ∠DEF\angle DEF∠DEF to ∠D′E′F′\angle D'E'F'∠D′E′F′. Their measures are 87∘87^\circ87∘ and (5v+2)∘(5v+2)^\circ(5v+2)∘. Find vvv.
Reflection preserves the angle measure at the corresponding vertex.
5v+2=875v+2=875v+2=87
5v=855v=855v=85
v=17v=17v=17
Reflection across mirror mmm sends PPP to P′P'P′. Their connector meets mmm perpendicularly at HHH. Given PH=2w+1PH=2w+1PH=2w+1 cm and HP′=15HP'=15HP′=15 cm, find www and PP′PP'PP′.
The mirror bisects the connector, so its halves have equal lengths.
2w+1=152w+1=152w+1=15
2w=142w=142w=14
w=7w=7w=7
The full connector contains both 15 cm halves.
PP′=15+15=30PP'=15+15=30PP′=15+15=30
A proposed reflection across y=xy=xy=x sends (2,−6)(2,-6)(2,−6) to (−6,−2)(-6,-2)(−6,−2). Which conclusion is correct?
Across y=xy=xy=x, exchange the coordinate roles while retaining each coordinate's sign.
(2,−6)↦(−6,2)(2,-6)\mapsto(-6,2)(2,−6)↦(−6,2)
The proposed second coordinate is -2 instead of 2, so that pair does not fit.
Which point lies on the falling diagonal mirror y=−xy=-xy=−x and is therefore fixed?
On this mirror, the height is the negative of the horizontal coordinate. For (−4,4)(-4,4)(−4,4), 4=−(−4)4=-(-4)4=−(−4).
(−4,4)↦(−4,4)(-4,4)\mapsto(-4,4)(−4,4)↦(−4,4)
Reflect the entire line x=5x=5x=5 across x=−1x=-1x=−1. What is its image?
Every point on the starting line has horizontal coordinate 5, and its height is unrestricted. The vertical-mirror rule gives
x′=2(−1)−5=−7.x'=2(-1)-5=-7.x′=2(−1)−5=−7.
Every height remains unchanged, so the image is the entire vertical line x=−7x=-7x=−7.
Segment CDCDCD has C(2,4)C(2,4)C(2,4) and D(−3,1)D(-3,1)D(−3,1). Reflect it across y=1y=1y=1. Which images and fixed-point statement are correct?
Keep each horizontal coordinate and reflect each height around 1. For C,
y′=2(1)−4=−2.y'=2(1)-4=-2.y′=2(1)−4=−2.
For D, its height already equals the mirror height, so it stays at (−3,1)(-3,1)(−3,1). Only D is fixed among the endpoints.
Reset this practice set?
This clears every answer you have given and starts the set again from question 1.