Question

Stephen draws figure ABCD in the coordinate plane

Answer 1

A=(-4,2)

B=(-2,2)

C=(-1,6)

D=(-5,6)

First, we need to translate it to the origin:

B=(-2,2)----------->(x+2,y-2)------------------>B'=(0,0)

A=(-4,2)----------->(x+2,y-2)------------------>A'=(2,0)

C=(-1,6)----------->(x+2,y-2)------------------>C'=(1,4)

D=(-5,6)----------->(x+2,y-2)------------------>D'=(-3,4)

Now, let's do the rotation using the following formula:

$\begin{gathered} x^{\prime}=x\cos (\theta)+y\sin (\theta) \\ y^{\prime}=-x\sin (\theta)+y\cos (\theta) \\ x^{\prime}=x\cos (270)+y\sin (270)=-y \\ y^{\prime}=-x\sin (270)+\cos (270)=x \end{gathered}$

So:

B'=(0,0)----------->(-y,x)------------------>B''=(0,0)

A'=(2,0)----------->(-y,x)------------------>A''=(0,2)

C'=(1,4)----------->(-y,x)------------------>C''=(-4,1)

D'=(-3,4)----------->(-y,x)------------------>D''=(-4,-3)

Now, translate the figure to its original position:

B''=(0,0)----------->(x-2,y+2)------------------>B'''=(-2,2)

A''=(0,2)----------->(x-2,y+2)------------------>A'''=(-2,4)

C''=(-4,1)----------->(x-2,y+2)------------------>C'''=(-6,3)

D''=(-4,-3)----------->(x-2,y+2)------------------>D'''=(-6,-1)