The coordinates of the vertices of the original shape are given to be:
[tex]\begin{gathered} L\Rightarrow(6,6) \\ M\Rightarrow(0,4) \\ N\Rightarrow(0,6) \end{gathered}[/tex]
The transformed shape coordinates are given to be:
[tex]\begin{gathered} L^{\prime}\Rightarrow(7,2) \\ M^{\prime}\Rightarrow(4,1) \\ N^{\prime}\Rightarrow(4,2) \end{gathered}[/tex]
On observation, the shapes show a dilation and a shift in position.
The image has a scale factor of 1/2, meaning it is reduced by 1/2. Therefore, the original coordinates will be reduced to give:
[tex]\begin{gathered} L^{\doubleprime}\Rightarrow\frac{1}{2}(6,6)=(3,3) \\ M^{\doubleprime}\Rightarrow\frac{1}{2}(0,4)=(0,2) \\ N^{\doubleprime}\Rightarrow(0,6)\Rightarrow(0,3) \end{gathered}[/tex]
The translation of the initial image can be gotten by subtracting the corresponding coordinates:
[tex]\begin{gathered} L^{\prime}-L^{\doubleprime}=(7-3,2-3)=4,-1 \\ M^{\prime}-M^{\doubleprime}=(4-0,1-2)=4,-1 \\ N^{\prime}-N^{\doubleprime}=(4-0,2-3)=4,-1 \end{gathered}[/tex]
All the differences are the same. This means that the image moves to the right by 4 units and down by 1 unit.
The dilation rule with a scale factor of k is given to be:
[tex](x,y)\to(kx,ky)[/tex]
The translation rule for a units to the right and b units down is given to be:
[tex](x,y)\to(x+a,y-b)[/tex]
Combining both rules, we have:
[tex](x,y)\to(kx+a,ky-b)[/tex]
Given:
[tex]\begin{gathered} k=\frac{1}{2} \\ a=4 \\ b=1 \end{gathered}[/tex]
Therefore, the transformation is given to be:
[tex]\Rightarrow(\frac{1}{2}x+4,\frac{1}{2}y-1)[/tex]
