Chapter Test
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Rotation:270^(∘) counterclockwise
Dilation: (a,b)→ (1/2a,1/2b)
Translation: (a,b)→ (a+1,b-1)
How many degrees do we have to rotate △ ABC to give it the same orientation as △ DEF? Let's compare the slopes of the corresponding sides BC and EF.
As we can see, - 1 and 1 are opposite reciprocals which means BC and EF are perpendicular to each other. This means, we can rotate △ ABC either 90^(∘) clockwise or 270^(∘) counterclockwise to give the triangles the same orientation. When we rotate a figure 270^(∘) about the origin, the vertices of the figure change in the following way. (a,b)→ (b,- a) Let's use this rule on the coordinates of the vertices of △ A'B'C'.
| Point | (a,b) | (b,- a) |
|---|---|---|
| A | (0,0) | (0,0) |
| B | (2,4) | (4,- 2) |
| C | (4,2) | (2,- 4) |
Now we can draw △ A'B'C' in the coordinate plane.
Next we will make sure the triangles have the same size. For this purpose, we have to calculate the length of two corresponding sides. Using the Distance Formula we can, for example, calculate the distance of EF and B'C'
| Segment | Points | sqrt((x_2-x_1)^2+(y_2-y_1)^2) | d |
|---|---|---|---|
| EF | ( 3,- 2) & ( 2,- 3) | sqrt(( 3- 2)^2+( - 2-( - 3))^2) | sqrt(2) |
| B'C' | ( 4,- 2) & ( 2,- 4) | sqrt(( 4- 2)^2+( - 2-( - 4))^2) | sqrt(8) |
By dividing EF with B'C' we can find which scale factor we have to use. Scale Factor: EF/B'C'=sqrt(2)/sqrt(8)=1/2 By dilating △ A'B'C' with a scale factor of 12, we can make sure the triangles have the same size.
| Point | (a,b) | (1/2a,1/2b) |
|---|---|---|
| A' | (0,0) | (0,0) |
| B' | (4,- 2) | (2,- 1) |
| C' | (2,- 4) | (1,- 2) |
Now we can draw △ A''B''C'' in the coordinate plane.
Finally, we have to translate △ A''B''C'' by moving one unit to the right and one unit down.
Now we can describe the similarity transformation that maps △ ABC onto △ CDB. Rotation:& 270^(∘) counterclockwise Dilation:& (a,b)→ (1/2a,1/2b) Translation:& (a,b)→ (a+1,b-1)
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