12 multiple-choice questions, progressively harder.
One translation sends A(−3,4)A(-3,4)A(−3,4) to A′(2,−2)A'(2,-2)A′(2,−2) and B(t,1)B(t,1)B(t,1) to B′(8,−5)B'(8,-5)B′(8,−5). Find ttt.
Solution
Correct answer: D
The first pair gives displacement (5,−6)(5,-6)(5,−6). The vertical coordinates of the second pair agree, since 1−6=−51-6=-51−6=−5.
Its horizontal coordinates must satisfy
t+5=8.t+5=8.t+5=8.
Therefore t=3t=3t=3.
A proposed translation maps A(1,1)A(1,1)A(1,1) to A′(4,−1)A'(4,-1)A′(4,−1), B(−2,0)B(-2,0)B(−2,0) to B′(1,−2)B'(1,-2)B′(1,−2), and C(0,3)C(0,3)C(0,3) to C′(3,2)C'(3,2)C′(3,2). Which statement is correct?
Correct answer: B
The first pair has change (3,−2)(3,-2)(3,−2). The second has change (3,−2)(3,-2)(3,−2) as well.
For the third, the horizontal change is 3−0=33-0=33−0=3. Its vertical change is
2−3=−1.2-3=-1.2−3=−1.
Its vertical change differs, so only the first two agree.
A translation sends A(−4,6)A(-4,6)A(−4,6) to itself. What must it do to B(5,−2)B(5,-2)B(5,−2)?
Correct answer: C
For A, both image-minus-start changes are zero. Every point has that same displacement.
(x,y)↦(x,y)(x,y)\mapsto(x,y)(x,y)↦(x,y)
This is the zero translation, so B also stays at (5,−2)(5,-2)(5,−2).
A translation pairs angles whose measures are 3p+43p+43p+4 degrees and 5p−185p-185p−18 degrees. Find their common measure.
Correct answer: A
Set the corresponding angle measures equal.
3p+4=5p−183p+4=5p-183p+4=5p−18
This gives 22=2p22=2p22=2p, so p=11p=11p=11. Substitute into either measure.
3(11)+4=373(11)+4=373(11)+4=37
The common measure is 37∘37^\circ37∘.
An unknown transformation agrees with one translation at three tested points. Is that enough to prove the unknown rule is that translation at every point in the plane?
The checks establish only what happened at the tested points. A transformation assigns an image to every point of the plane, and its rule could change an untested point differently. A specified translation rule determines all images; agreement at a few samples does not establish an arbitrary unknown rule everywhere.
Point PPP lies between AAA and BBB, with AP=4AP=4AP=4 cm and PB=9PB=9PB=9 cm. A translation maps them to A′,P′,B′A',P',B'A′,P′,B′. Which pair gives A′P′A'P'A′P′ and A′B′A'B'A′B′?
The original segment length is found by segment addition.
AB=4+9=13 cmAB=4+9=13\text{ cm}AB=4+9=13 cm
The translation preserves both AP and AB. Thus A′P′=4A'P'=4A′P′=4 cm and A′B′=13A'B'=13A′B′=13 cm.
A translation has displacement (−6,2)(-6,2)(−6,2). The image of B(s,s−9)B(s,s-9)B(s,s−9) is B′(0,−1)B'(0,-1)B′(0,−1). What is the preimage point BBB?
The horizontal image coordinate requires
s−6=0.s-6=0.s−6=0.
So s=6s=6s=6, and the original vertical coordinate is 6−9=−36-9=-36−9=−3. Check that adding 2 gives the required image height −1-1−1. Thus B=(6,−3)B=(6,-3)B=(6,−3).
In the specified correspondence, A(0,0)↦A′(0,0)A(0,0)\mapsto A'(0,0)A(0,0)↦A′(0,0) and B(2,0)↦B′(0,2)B(2,0)\mapsto B'(0,2)B(2,0)↦B′(0,2). Both segments have length 2. Can a translation make these matches?
A has displacement (0,0)(0,0)(0,0), so a translation making that match would leave every point in place.
B instead changes its horizontal coordinate by
0−2=−2,0-2=-2,0−2=−2,
and its vertical coordinate by
2−0=2.2-0=2.2−0=2.
Its displacement is (−2,2)(-2,2)(−2,2). The two point displacements disagree, so the specified correspondence is not a translation. Equal segment lengths alone do not establish a translation.
A rule moves every point eight units right and six units down. Which statement about its displacement is justified here?
The displacement pair records two coordinate changes.
(8,−6)(8,-6)(8,−6)
The straight arrow from a point to its image is diagonal. The horizontal and vertical counts describe separate directions, so their sum is not its straight length. All points still share the same arrow direction and length.
A translation has rule (x,y)↦(x+3,y)(x,y)\mapsto(x+3,y)(x,y)↦(x+3,y). What is the image of the entire line y=2y=2y=2?
Every point on the line has form (x,2)(x,2)(x,2) and maps to (x+3,2)(x+3,2)(x+3,2). Its image height is
y′=2.y'=2.y′=2.
Every point (u,2)(u,2)(u,2) is reached from (u−3,2)(u-3,2)(u−3,2), so the image covers that whole line. The line is unchanged as a set, although its individual points move.
A translation sends A(−1,2)A(-1,2)A(−1,2) to A′(4,−2)A'(4,-2)A′(4,−2). Segments ABABAB and CDCDCD are parallel. Which conclusion is guaranteed without knowing the other coordinates?
The displacement is (5,−4)(5,-4)(5,−4), but its values are not needed for the relationship. Applying the same slide preserves the directions of both parallel segments, so A′B′∥C′D′A'B'\parallel C'D'A′B′∥C′D′. Parallel segments need not have equal lengths.
A translation sends A(u,1)A(u,1)A(u,1) to A′(7,4)A'(7,4)A′(7,4) and B(−2,v)B(-2,v)B(−2,v) to B′(3,1)B'(3,1)B′(3,1). Which displacement and missing values fit both pairs?
The B pair determines the horizontal change.
3−(−2)=53-(-2)=53−(−2)=5
The A pair determines the vertical change.
4−1=34-1=34−1=3
Now u+5=7u+5=7u+5=7 gives u=2u=2u=2, and v+3=1v+3=1v+3=1 gives v=−2v=-2v=−2. Substitution checks both specified images.
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