Question

A parallelogram has coordinates A(1, 1), B(5, 4), C(7, 1), and D(3, -2). What are the coordinates of parallelogram A′B′C′D′ after a 180° rotation about the origin and a translation 5 units to the right and 1 unit down?

Answer 1

Answer:

$A'(4,-2)$ $B'(0,-5)$ $C'(-2,-2)$ $D'(2,1)$

Step-by-step explanation:

The given parallelogram has coordinates A(1, 1), B(5, 4), C(7, 1), and D(3, -2).

The rotation of $180\degree$about the origin  has the mapping;

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

This implies that;

$A(1,1)\rightarrow (-1,-1)$

$B(5,4)\rightarrow (-5,-4)$

$C(7,1)\rightarrow (-7,-1)$

$D(3,-2)\rightarrow (-3,2)$

A translation of  5 units to the right and 1 unit down has the mapping;

$P(x,y)\rightarrow P'(x+5,y-1)$

We apply this to the resulting coordinates to obtain;

$A(1,1)\rightarrow (-1,-1) \rightarrow (-1+5,-1-1)=A'(4,-2)$

$B(5,4)\rightarrow (-5,-1) \rightarrow (-5+5,-4-1)=B'(0,-5)$

$C(7,1)\rightarrow (-7,-1)\rightarrow (-7+5,-1-1)=C'(-2,-2)$

$D(3,-2)\rightarrow (-3,2)\rightarrow (-3+5,2-1)=D'(2,1)$