• Given the line AB whose endpoints are:
[tex]\begin{gathered} A\mleft(4,4\mright) \\ B\mleft(5,-1\mright) \end{gathered}[/tex]
You know that the line of reflection is:
[tex]y=x[/tex]
By definition, the rule for that Reflection is:
[tex](x,y)\rightarrow\mleft(y,x\mright)[/tex]
Then, the endpoints of the Image A'B' are:
[tex]\begin{gathered} A(4,4)\rightarrow A^{\prime}(4,4) \\ B(5,-1)\rightarrow B^{\prime}(-1,5) \end{gathered}[/tex]
Therefore, you can notice that the lines of AB and A'B' are graphed correctly.
• According to the information given in the exercise, you must find the perimeter of the figure formed by these segments:
[tex]\begin{gathered} AB \\ B^{\prime}B \\ AB^{\prime} \end{gathered}[/tex]
Look at the following picture, where you can see the segments that form a triangle:
You can see in the picture a triangle that is formed by the segments AB, B’B and AB'.
To find the length of each segment by using the formula for calculating the distance between two points:
[tex]d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Where the points are:
[tex](x_1,y_1);(x_2,y_2)_{}[/tex]
Therefore, knowing the endpoints of each segment of the triangle, you get:
- Length of the side AB:
[tex]AB=\sqrt[]{(5-4)^2+(-1-4)^2}[/tex][tex]\begin{gathered} AB=\sqrt[]{(1)^2+(-5)^2} \\ AB=\sqrt[]{26} \end{gathered}[/tex]
- Length of the side AB':
[tex]A^{}B^{\prime}=\sqrt[]{(-1-4)^2+(5-4)^2}=\sqrt[]{26}[/tex]
- Length of the side B'B:
[tex]B^{\prime}B=\sqrt[]{(-1-5)^2+(5-(-1))^2}=6\sqrt[]{2}[/tex]
Knowing the length of each side, you can add them in order to find the perimeter:
[tex]P=\sqrt[]{26}+\sqrt[]{26}+6\sqrt[]{2}\approx18.7\text{ }units[/tex]
Therefore, the answer is:
[tex]P\approx18.7\text{ }units[/tex]