12 multiple-choice questions, progressively harder.
Reflection across y=xy=xy=x sends P(2t−1,8)P(2t-1,8)P(2t−1,8) to P′(8,5)P'(8,5)P′(8,5). Find ttt.
Solution
Correct answer: D
The image's second coordinate equals the starting first coordinate.
2t−1=52t-1=52t−1=5
2t=62t=62t=6
t=3t=3t=3
The starting point is (5,8)(5,8)(5,8), which swaps to (8,5)(8,5)(8,5).
Can reflection across x=2x=2x=2 make both matches A(−1,4)↦A′(5,4)A(-1,4)\mapsto A'(5,4)A(−1,4)↦A′(5,4) and B(7,−3)↦B′(−3,−3)B(7,-3)\mapsto B'(-3,-3)B(7,−3)↦B′(−3,−3)?
Correct answer: B
For A, the mirror rule gives
2(2)−(−1)=5.2(2)-(-1)=5.2(2)−(−1)=5.
For B, it gives
2(2)−7=−3.2(2)-7=-3.2(2)−7=−3.
Each height stays unchanged. Both pairs satisfy the same reflection rule, even though their travel distances differ.
The points P(−3,1)P(-3,1)P(−3,1) and Q(3,5)Q(3,5)Q(3,5) are equally far from the yyy-axis on opposite sides. Is QQQ the reflection of PPP across that axis?
Correct answer: C
Reflection across the y-axis keeps the height. P must map to (3,1)(3,1)(3,1), not (3,5)(3,5)(3,5).
The proposed connector is oblique rather than horizontal, so it fails the perpendicular condition.
A proposed reflection sends A(1,2)A(1,2)A(1,2) to A′(5,2)A'(5,2)A′(5,2) and B(2,4)B(2,4)B(2,4) to B′(4,5)B'(4,5)B′(4,5). Which statement decides whether one reflection can do both?
Correct answer: A
AA-prime is horizontal, so its perpendicular bisector is the vertical line through its midpoint.
h=1+52=3h=\frac{1+5}{2}=3h=21+5=3
Reflection across x=3x=3x=3 sends B to (4,4)(4,4)(4,4). It keeps height 4, so it cannot produce the proposed height 5.
Triangle ABCABCABC has A(−2,1)A(-2,1)A(−2,1), B(3,1)B(3,1)B(3,1), and C(0,4)C(0,4)C(0,4). Reflect it across y=−xy=-xy=−x. Which side corresponds to ABABAB, and what is its length?
The falling-diagonal rule sends A to (−1,2)(-1,2)(−1,2) and B to (−1,−3)(-1,-3)(−1,−3). Their images are the endpoints of the corresponding side A′B′A'B'A′B′.
The original side is horizontal with length 3−(−2)=53-(-2)=53−(−2)=5. The image side is vertical with the same length, also checked by
2−(−3)=5.2-(-3)=5.2−(−3)=5.
For what value of ttt is P(t,3t−8)P(t,3t-8)P(t,3t−8) fixed by reflection across y=xy=xy=x?
A fixed point lies on the mirror, so its two coordinates must agree.
t=3t−8t=3t-8t=3t−8
2t=82t=82t=8
t=4t=4t=4
This gives (4,4)(4,4)(4,4) on y=xy=xy=x.
Reflection across y=−2y=-2y=−2 sends the entire horizontal line y=5y=5y=5 to which line, and does it fix any point of y=5y=5y=5?
Every starting point has height 5. Reflecting this height around -2 gives
y′=2(−2)−5=−9.y'=2(-2)-5=-9.y′=2(−2)−5=−9.
The horizontal coordinate ranges over all real numbers, so the image is y=−9y=-9y=−9. The starting line has no point on the mirror y=−2y=-2y=−2, so it has no fixed point.
A line passes through the midpoint of a nonzero point-image connector, but meets it at a 60-degree angle. Can it be the mirror for that pair?
The midpoint condition gives equal halves, but the reflection definition also requires a right angle. The stated 60-degree intersection fails that second requirement.
A reflection in a vertical line sends P(11,−4)P(11,-4)P(11,−4) to P′(−5,−4)P'(-5,-4)P′(−5,−4). What image does it give Q(−2,9)Q(-2,9)Q(−2,9)?
Locate the mirror halfway between the first pair's horizontal coordinates.
h=11+(−5)2=3h=\frac{11+(-5)}{2}=3h=211+(−5)=3
Reflect Q across x=3x=3x=3, keeping its height.
x′=2(3)−(−2)=8x'=2(3)-(-2)=8x′=2(3)−(−2)=8
Thus Q′(8,9)Q'(8,9)Q′(8,9).
Distinct points AAA and BBB are fixed by a reflection. Point CCC is not on line ABABAB. Which mirror and image behavior are forced?
Every fixed point lies on the mirror, so both A and B lie on it. Two distinct points determine exactly one line, making AB the mirror. C is off that line, so it maps to the opposite side and is not fixed.
A reflection maps one angle of measure (6n+5)∘(6n+5)^\circ(6n+5)∘ to its corresponding angle of measure (4n+29)∘(4n+29)^\circ(4n+29)∘. Find their common measure.
Corresponding angle measures are equal.
6n+5=4n+296n+5=4n+296n+5=4n+29
2n=242n=242n=24
n=12n=12n=12
Substitute into the first measure.
6(12)+5=776(12)+5=776(12)+5=77
The other expression also gives 77 degrees.
You have drawn the perpendicular from an off-mirror point VVV to an oblique mirror, meeting it at HHH. Which next step places the image of VVV correctly?
The perpendicular supplies the connector line. The image must be on the opposite side of the mirror with equal distance to the foot. Extending through H and copying HV onto the opposite ray supplies both remaining conditions.
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