Answer:
ΔRST ≅ ΔBAC
Step-by-step explanation:
Consider triangles ABC with vertices A(-3,-1), B(-1,2) and C(-5,3) and RST with vertices R(1,1), S(3,4) and T(5,0).
The rotation by 180° about the origin has the rule
[tex](x,y)\rightarrow (-x,-y)[/tex]
So,
- [tex]R(1,1)\rightarrow R'(-1,-1);[/tex]
- [tex]S'(3,4)\rightarrow S'(-3,-4);[/tex]
- [tex]T(5,0)\rightarrow T'(-5,0).[/tex]
Translation 3 units up has the rule
[tex](x,y)\rightarrow (x,y+3)[/tex]
Hence
- [tex]R'(-1,-1)\rightarrow (-1,2)[/tex] that is exactly point B;
- [tex]S'(-3,-4)\rightarrow (-3,-1)[/tex] that is exactly point A;
- [tex]T'(-5,0)\rightarrow (-5,3)[/tex] that is exactly points C.
Therefore, triangle RST is congruent to BAC.
