Question

Indicate the transformation that has occurred.2.A. (x,y)-->(-x+3.y-5) C. (x,y) --> (-x,y-5)B. (x,y) --> (x +3,y-5) D. (x,y) --> (x-1,-y)

Answer 1

So we have a transformation that maps a triangle into another one. This is made by transforming the points X, Y and Z into X', Y' and Z'. In order to find out which of the four options is the correct one we must verify that points X, Y and Z actually transform into X', Y' and Z'.

We have:

`X=(2,5)\rightarrow X^(\prime)=(1,0)`

Let's see which of the four transformations do this. So for A:

`\begin{gathered} (x,y)\rightarrow(-x+3,y-5) \\ X=(2,5)=(x,y) \\ \text{Then} \\ X^(\prime)=(-x+3,y-5)=(-2+3,5-5) \\ X^(\prime)=(1,0) \end{gathered}`

So transformation A is a possible answer, let's see the rest.

For C:

`\begin{gathered} (x,y)\rightarrow(-x,y-5) \\ X=(2,5)=(x,y) \\ \text{Then} \\ X^(\prime)=(-x,y-5)=(-2,5-5) \\ X^(\prime)=(-2,0)\\e(1,0) \end{gathered}`

So the X' that we calculate with transformation C is different that the one we are looking for so we discard this option.

For option B we have:

`\begin{gathered} (x,y)\rightarrow(x+3,y-5) \\ X=(2,5)=(x,y) \\ \text{Then} \\ X^(\prime)=(x+3,y-5)=(2+3,5-5)=(5,0) \\ X^(\prime)=(5,0)\\e(1,0) \end{gathered}`

Like what happened with C, transformation B is discarded.

Let's see what happens with D:

`\begin{gathered} (x,y)\rightarrow(x-1,-y) \\ X=(2,5)=(x,y) \\ \text{Then} \\ X^(\prime)=(x-1,-y)=(2-1,-5)=(1,-5) \\ X^(\prime)=(1,-5)=(1,0) \end{gathered}`

So D is also discarded. This would mean that A is the correct option but just in case, let's check if it tansform points Y=(0,2) and Z=(3,1) into Y'=(3,-3) and Z'=(0,-4):

`\begin{gathered} (x,y)\rightarrow(-x+3,y-5) \\ \text{If} \\ Y=\mleft(0,2\mright) \\ \text{Then} \\ Y^(\prime)=(-0+3,2-5)=(3,-3) \\ \text{If} \\ Z=\mleft(3,1\mright) \\ \text{Then} \\ Z^(\prime)=(-3+3,1-5)=(0,-4) \end{gathered}`

So Y' and Z' are (3,-3) and (0,-4) which definetely means that option A is the correct one.