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Can you spot any symmetries?
Example Solution: A reflection in the line x=-1 followed by a dilation centered at R(-3,0) and a scale factor of 3.
To describe a similarity transformation that maps â–ł ABC to â–ł RST, let's begin by plotting the two triangles.
We can see that the reflection in the line x=-1 will line up figures' orientations. Let's do it then, remembering that each vertex must be moved to the opposite side of the line of reflection while maintaining the same distance from the line.
The reflection mapped â–ł A'B'C' completely onto â–ł RST. This indicates that â–ł RST might be a dilation of â–ł A'B'C' with respect to the common vertex R. We can check that by comparing the ratios of the lengths of the corresponding sides. To find the lengths, let's use the Distance Formula.
Side | Points | sqrt((x_2-x_1)^2+(y_2-y_1)^2) | Length |
---|---|---|---|
RC' | ( - 3,0) ( - 1,-2) | sqrt(( -1-( - 3))^2+( -2- 0)^2) | sqrt(8) |
RB' | ( -3,0) ( 0,-1) | sqrt(( 0-( -3))^2+( -1- 0)^2) | sqrt(10) |
B'C' | ( 0,-1) ( -1,-2) | sqrt(( -1- 0)^2+( -2-( -1))^2) | sqrt(2) |
RT | ( -3,0) ( 3,-6) | sqrt(( 3-( -3))^2+( -6- 0)^2) | sqrt(72) |
RS | ( -3,0) ( 6,- 3) | sqrt(( 6-( -3))^2+( -3- 0)^2) | sqrt(90) |
ST | ( 6,-3) ( 3,-6) | sqrt(( 3- 6)^2+( -6-( -3))^2) | sqrt(18) |
Once we know the lengths, let's compare them. In our case the preimage is â–ł A'B'C' and the image is â–ł RST.
Side | Corresponding side | Ratio |
---|---|---|
RC' | RT | sqrt(72)/sqrt(8)=3 |
RB' | RS | sqrt(90)/sqrt(10)=3 |
B'C' | ST | sqrt(18)/sqrt(2)=3 |
Since all the ratios are equal 3, we know that â–ł RST is a dilation of â–ł A'B'C' with respect to R and a scale factor 3. Therefore, the similarity transformation that maps â–ł ABC to â–ł RST is a reflection in the line x=-1 followed by a dilation with center R(-3,0) and scale factor 3.