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:
[tex]X=(2,5)\rightarrow X^{\prime}=(1,0)[/tex]Let's see which of the four transformations do this. So for A:
[tex]\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}[/tex]So transformation A is a possible answer, let's see the rest.
For C:
[tex]\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)\ne(1,0) \end{gathered}[/tex]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:
[tex]\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)\ne(1,0) \end{gathered}[/tex]Like what happened with C, transformation B is discarded.
Let's see what happens with D:
[tex]\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}[/tex]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):
[tex]\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}[/tex]So Y' and Z' are (3,-3) and (0,-4) which definetely means that option A is the correct one.
