QUESTION A
ΔALT has coordinates [tex]A(-5,-1),L(-3,-2),T(-3,2)[/tex].
If we translate by the rule [tex](x,y)\rightarrow (x+6,y-3)[/tex]
Then;
[tex]A(-5,-1)\rightarrow (-5+6,-1-3)=(1,-4)[/tex].
[tex]B(-3,-2)\rightarrow (-3+6,-2-3)=(3,-5)[/tex].
[tex]T(-3,2)\rightarrow (-3+6,2-3)=(3,-1)[/tex].
A reflection over the y-axis has the mapping
[tex](x,y)\rightarrow (-x,y)[/tex]
If we reflect the resulting points over the y-axis, we obtain,
[tex]A(-5,-1)\rightarrow (1,-4)\rightarrow A'(-1,-4)[/tex].
[tex]B(-3,-2)\rightarrow (3,-5)\rightarrow B'(-3,-5)[/tex].
[tex]T(-3,2)\rightarrow (3,-1)\rightarrow C'(-3,-1)[/tex].
QUESTION B
ΔTAB has vertices [tex]T(2,3),A(1,1),B(4,-3)[/tex].
A reflection over the x-axis has the mapping
[tex](x,y)\rightarrow (x,-y)[/tex]
A reflection over the y-axis also has the mapping;
[tex](x,y)\rightarrow (-x,y)[/tex]
If we reflect ΔTAB over the x-axis and reflect the resulting image over the y-axis, we obtain;
[tex]T(2,3)\rightarrow (2,-3) \rightarrow T'(-2,-3)[/tex]
[tex]A(1,1)\rightarrow (1,-1) \rightarrow A'(-1,-1)[/tex]
[tex]B(4,-3)\rightarrow (4,3) \rightarrow B'(-4,3)[/tex]
QUESTION C
ΔALT has vertices [tex]A(-5,-1),L(-3,-2),T(-3,2)[/tex].
A [tex]90\degree[/tex] clockwise rotation about the origin has the mapping;
[tex](x,y)\rightarrow (y,-x)[/tex]
A reflection in the line y=x also has the mapping;
[tex](x,y)\rightarrow (y,x)[/tex]
When we rotate the given triangle [tex]90\degree[/tex] clockwise about the origin, and then reflect the image over the line y=x, we obtain;
[tex]A(-5,-1)\rightarrow (-1,5)\rightarrow A'(5,-1)[/tex]
[tex]L(-3,-2)\rightarrow (-2,3)\rightarrow L'(3,-2)[/tex]
[tex]T(-3,2)\rightarrow (2,3)\rightarrow T'(3,2)[/tex]
QUESTION D
ΔTAB has vertices [tex]T(2,3),A(1,1),B(4,-3)[/tex].
A reflection over the y-axis has the mapping;
[tex](x,y)\rightarrow (-x,y)[/tex]
The rule for the given translation is
[tex](x,y)\rightarrow (x+2,y-1)[/tex]
If reflect the given triangle over the y-axis and translate using the rule [tex](x,y)\rightarrow (x+2,y-1)[/tex], we obtain;
[tex]T(2,3)\rightarrow (-2,3)\rightarrow (-2+2,3-1)=T'(0,2)[/tex]
[tex]A(1,1)\rightarrow (-1,1)\rightarrow (-1+2,1-1)=A'(1,0)[/tex]
[tex]B(4,-3)\rightarrow (-4,-3)\rightarrow (-4+2,-3-1)=B'(-2,-4)[/tex]
