Question

square a\b\c\d\ is the final image after the rule $t_{ - 4,-1}circ r_{0,90^{circ}}(x,y)$ was applied to square abcd. what are the coordinates of vertex a of square abcd? (-1, -6) (-1, -2) (-1, 6) (-2, 1)

Explanation:

1. First, understand the transformation rules:

• The transformation rule \(T_{ - 4,-1}\circ R_{0,90^{\circ}}(x,y)\) means a \(90 - \)degree counter - clockwise rotation about the origin (\(R_{0,90^{\circ}}\)) followed by a translation of \((-4,-1)\). The rule for a \(90^{\circ}\) counter - clockwise rotation about the origin is \(R_{0,90^{\circ}}(x,y)=(-y,x)\), and the rule for a translation \(T_{a,b}(x,y)=(x + a,y + b)\) where \(a=-4\) and \(b = - 1\). So the combined transformation rule is \(T_{ - 4,-1}(R_{0,90^{\circ}}(x,y))=T_{ - 4,-1}(-y,x)=(-y-4,x - 1)\).

1. Let the coordinates of \(A\) in square \(ABCD\) be \((x,y)\). After the transformation, assume the coordinates of \(A''\) (the final image of \(A\)) are \((x_{A''},y_{A''})\).

• From the graph, assume the coordinates of \(A''\) are \((-5, - 3)\).
• We know that \(x_{A''}=-y - 4\) and \(y_{A''}=x - 1\).
• Substitute \(x_{A''}=-5\) and \(y_{A''}=-3\) into the equations:
• For \(x_{A''}=-y - 4\), we have \(-5=-y - 4\). Solving for \(y\):
• Add 4 to both sides of the equation: \(-5 + 4=-y\), so \(y = 1\).
• For \(y_{A''}=x - 1\), we have \(-3=x - 1\). Solving for \(x\):
• Add 1 to both sides of the equation: \(x=-3 + 1=-2\).

1. So the coordinates of vertex \(A\) of square \(ABCD\) are \((-2,1)\).

Answer:

D. \((-2,1)\)