To describe a that maps â–³ ABC to â–³ RST, let's begin by plotting the triangles with the given vertices.
Examining the two triangles, we can see that they have different orientations. In △ ABC, vertex B lies on the y-axis and in △ MNP vertex S lies on the x-axis. Therefore, if we rotate △ ABC 270^(∘) about the origin, we will line up their orientations. To perform the rotation, we can use the coordinate rule.
preimage (a,b) → image (b,- a)When a point (a,b) is rotated 270^(∘) about the origin, its image has coordinates (b,- a). With this in mind, we can find coordinates of the image.
| (a,b)
|
(b,- a)
|
| A(3,- 2)
|
A'(- 2,-3)
|
| B(0,4)
|
B'(4,0)
|
| C(-1,-3)
|
C'(-3,1)
|
Let's plot the image â–³ A'B'C'.
Now the figures have the same orientation, but they still differ in size. Therefore, we need to perform a . To find the scale factor, we should compare two corresponding sides. Let's choose the shortest ones: A'C' and RT. We can find their lengths by using the .
| Side
|
Points
|
sqrt((x_2-x_1)^2+(y_2-y_1)^2)
|
d
|
| A'C'
|
( -2,-3) ( -3,1)
|
sqrt(( -3-( -2))^2+( 1-( -3))^2)
|
sqrt(17)
|
| RT
|
( -4,-6) ( -6,2)
|
sqrt(( -6-( -4))^2+( -2-( -6))^2)
|
sqrt(68)
|
Once we know the lengths, let's compare them. In our case the preimage is â–³ A'B'C' and the image is â–³ RST.
k=RT/A'C'=sqrt(68)/sqrt(17)
k=sqrt(68)/sqrt(17)
k=sqrt(68/17)
k=sqrt(4/1)
k=sqrt(4)
k=2
Therefore, if we dilate â–³ A'B'C using a scale factor of 2, the triangles will have the same size. To perform this dilation, we should multiply preimage's coordinates by 2.
| (a,b)
|
(2a,2b)
|
| A'(-2,-3)
|
A''(-4,-6)
|
| B'(4,0)
|
B''(8,0)
|
| C'(-3,1)
|
C''(-6,2)
|
Now we have everything needed to plot the image â–³ A''B''C''.
Finally, the vertices of △ A''B''C'' map onto △ RST. Therefore, the similarity transformation that maps △ ABC to △ RST is a 270^(∘) rotation about the origin followed by a dilation with a scale factor of 2.
