12 multiple-choice questions, progressively harder.
Triangle ABCABCABC has A(−2,−1)A(-2,-1)A(−2,−1), B(1,−1)B(1,-1)B(1,−1), and C(−1,2)C(-1,2)C(−1,2). Translate it two units right and one unit up. Which image is correct?
Solution
Correct answer: B
The displacement pair is (2,1)(2,1)(2,1). Add it to every vertex, not just A.
For A, adding the changes gives
A′=(0,0).A'=(0,0).A′=(0,0).
For B, the image is (1+2,−1+1)=(3,0)(1+2,-1+1)=(3,0)(1+2,−1+1)=(3,0).
For C, the image is (−1+2,2+1)=(1,3)(-1+2,2+1)=(1,3)(−1+2,2+1)=(1,3).
Under (x,y)↦(x−5,y+2)(x,y)\mapsto(x-5,y+2)(x,y)↦(x−5,y+2), the image is P′(−3,7)P'(-3,7)P′(−3,7). What was PPP?
Correct answer: D
The starting horizontal coordinate must become -3 after subtracting 5.
x−5=−3x-5=-3x−5=−3
So x=2x=2x=2. The vertical coordinate must satisfy y+2=7y+2=7y+2=7, so y=5y=5y=5. Substituting (2,5)(2,5)(2,5) into the rule gives (−3,7)(-3,7)(−3,7).
A translation sends A(−5,1)A(-5,1)A(−5,1) to A′(−2,−3)A'(-2,-3)A′(−2,−3). Which second pair fits it?
Correct answer: A
The first pair gives changes −2−(−5)=3-2-(-5)=3−2−(−5)=3 and −3−1=−4-3-1=-4−3−1=−4.
For B, apply three right and four down.
x′=2+3=5x'=2+3=5x′=2+3=5
y′=6−4=2y'=6-4=2y′=6−4=2
Thus B′(5,2)B'(5,2)B′(5,2) fits.
A segment has endpoints U(−3,2)U(-3,2)U(−3,2) and V(2,2)V(2,2)V(2,2). A translation sends UUU to U′(1,−1)U'(1,-1)U′(1,−1). Where is V′V'V′?
Correct answer: C
The displacement from U to U-prime is (4,−3)(4,-3)(4,−3). Apply it to V.
x′=2+4=6x'=2+4=6x′=2+4=6
y′=2−3=−1y'=2-3=-1y′=2−3=−1
The image is V′(6,−1)V'(6,-1)V′(6,−1).
A translation sends A(7,−2)A(7,-2)A(7,−2) to A′(−1,3)A'(-1,3)A′(−1,3). What is the image of B(2,4)B(2,4)B(2,4)?
The changes are −1−7=−8-1-7=-8−1−7=−8 horizontally and 3−(−2)=53-(-2)=53−(−2)=5 vertically.
Apply them to B.
x′=2−8=−6x'=2-8=-6x′=2−8=−6
y′=4+5=9y'=4+5=9y′=4+5=9
So B′=(−6,9)B'=(-6,9)B′=(−6,9).
A translation maps segment ABABAB onto A′B′A'B'A′B′. Given AB=16AB=16AB=16 and A′B′=3s−2A'B'=3s-2A′B′=3s−2, find sss.
Corresponding lengths are equal.
3s−2=163s-2=163s−2=16
Add 2, then divide by 3.
3s=183s=183s=18
s=6s=6s=6
A translation sends angle JKLJKLJKL to angle J′K′L′J'K'L'J′K′L′. Their measures are 4w+74w+74w+7 degrees and 71∘71^\circ71∘. Find www.
Angle measure is preserved.
4w+7=714w+7=714w+7=71
Subtract 7 and divide by 4.
4w=644w=644w=64
w=16w=16w=16
The specified pairs are A(−2,−3)↦A′(0,2)A(-2,-3)\mapsto A'(0,2)A(−2,−3)↦A′(0,2) and B(4,1)↦B′(6,7)B(4,1)\mapsto B'(6,7)B(4,1)↦B′(6,7). Why do they fail to fit one translation?
For A the horizontal change is 2, and its vertical change is
2−(−3)=5.2-(-3)=5.2−(−3)=5.
For B the horizontal change is also 2, but its vertical change is
7−1=6.7-1=6.7−1=6.
The vertical changes disagree, so no single displacement pair fits both.
Apply displacement (−12,32)(-\tfrac12,\tfrac32)(−21,23) to T(52,−1)T(\tfrac52,-1)T(25,−1). What is T′T'T′?
Add the fractional changes to the matching coordinates.
x′=52−12=2x'=\frac52-\frac12=2x′=25−21=2
y′=−1+32=12y'=-1+\frac32=\frac12y′=−1+23=21
So T′=(2,12)T'=(2,\tfrac12)T′=(2,21).
A triangle side is 6 cm long. Each vertex travels 8 cm under a translation. Which statement is correct?
The travel distance describes the change in position of each point. It does not get added to a side length. Distance preservation keeps the image side 6 cm long, while both endpoints make the same 8 cm journey.
A translation maps X,Y,ZX,Y,ZX,Y,Z to X′,Y′,Z′X',Y',Z'X′,Y′,Z′ respectively. Which angle must have the same measure as ∠XYZ\angle XYZ∠XYZ?
The vertex Y maps to Y-prime. Ray YX maps to Y-prime X-prime, and ray YZ maps to Y-prime Z-prime. Those image rays form ∠X′Y′Z′\angle X'Y'Z'∠X′Y′Z′, whose measure is preserved.
A point P(r,2r)P(r,2r)P(r,2r) maps to P′(7,5)P'(7,5)P′(7,5) under displacement (3,−3)(3,-3)(3,−3). What is rrr?
The horizontal coordinate gives
r+3=7.r+3=7.r+3=7.
Thus r=4r=4r=4. Check the vertical coordinate with that value.
2(4)−3=52(4)-3=52(4)−3=5
Both coordinates agree with the specified image.
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