Question

△DEF is reflected to form​​ ​ △D′E′F′ ​. The vertices of ​ ​ △DEF ​ ​ are ​ D(−1,−1) ​ , ​ E′(1,−1) ​ , and ​ F(−1,−6) ​ . The vertices of ​ △P′Q′R′ ​ are D′(−1,1) , E′(1,1) , and F′(−1,6) . Which reflection results in the transformation of ​ ​ ​ △DEF ​ ​ ​​ to ​ ​ ​ △D′E′F′ ​ ​ ​​? reflection across the x-axis reflection across the y-axis reflection across y = x reflection across y=−x

Answer 1

Reflection across the x-axis because all y-coordinates for triangle DEF are the opposite for the y-coordinates in triangle D'E'F'.

Answer 2

Answer:

Triangle DEF is reflected across x -axis.

Step-by-step explanation:

Since, the rule of reflection across x -axis,

$(x,y)\rightarrow (x,-y)$

The rule of reflection across y -axis,

$(x,y)\rightarrow (-x,y)$

The rule of reflection across line y = x,

$(x,y)\rightarrow (y,x)$

The rule of reflection across y = -x ,

$(x,y)\rightarrow (-y,-x)$

Since, in transformation of the triangle DEF with vertices D(-1,-1) , ​E(1,-1) and ​ F(-1,-6) into the triangle D′E′F' with vertices D′(-1,1) , E′(1,1) , and F′(-1,6).

The rule of reflection is,

$(x,y)\rightarrow (x,-y)$

Thus, the triangle DEF is reflected across x -axis.