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Eva8 [605]
3 years ago
13

The following triangle CAT is transformed using a translation: Triangle CAT is shown. C is at 0, 0, A is at negative 2, 1, T is

at negative 1, 4. Which could be coordinates of C'A'T' after the translation? (4 points) Select one: a. C=(0,0), A=(2,1), T=(1,4) b. C=(2,0), A=(0,1), T=(1,4) c. C=(0,0), A=(−1,−2), T=(−4,−1) d. C=(−2,0), A=(−1,0), T=(−4,1)
Mathematics
2 answers:
jok3333 [9.3K]3 years ago
8 0

Solution: If a shape is being translated neither it's shape nor size changes.When we translate a shape we just move a shape or geometrical figure by certain distance either horizontally or vertically or both horizontally and vertically.

Coordinates of Triangle CAT is →C(0,0), and A(-2,1), and T(-1,4).

In option (a) vertices C has not been translated , but A and T are translated.Which is Untrue because we have to move all the vertices by same value. But x and y can have different value.

In option (b) all the coordinates C,A,and T is being translated along→ x by 2 units , horizontally right, but not translated in vertical direction.  

C(0,0)⇒C(0+2,0+0)→C(2,0)

A(-2,1)⇒C(-2+2,1+0)→C(0,1)

T(-1,4)⇒C(-1+2,4+0)→C(1,4)

In option C, Coordinates of A and B are changed but not of C, So in this case translation can't happen.

In option D , C is being translated 2 units horizontally left ,  but amount of translation in A and T are different.

So , Option (b) is true.

zysi [14]3 years ago
7 0

the answer would be b. C=(2,0), A=(0,1), T=(1,4)

all x's move 2 to the right and no change in y


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‹1, -1, 1› × ‹0, 1, 1›

You are interested in vectors orthogonal to the originals, which don't change when you scale them. Using 0,-1,1 is much easier than 6s and 7s.

So what methods are there to compute this? You can review them here (or presumably in your class notes or textbook):
http://en.wikipedia.org/wiki/Cross_produ...

In addition to these methods, sometimes I like to set up:
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That is the dot product, and having these dot products equal zero guarantees orthogonality. You can convert that to:

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This is two equations, three unknowns, so you can solve it with one free parameter:

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And take the negative for the other:

‹ 2/√6 , 1/√6 , -1/√6 ›
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