Question

the rule ( r_{0,90^{circ}}circ t_{-1,1}(x,y) ) is applied to ( \triangle bcd ) to produce ( \triangle bcd ). point ( b ) of the final image is at ( (-4,1) ). what are the coordinates of point ( b ) on the pre - image? ( (-2,-5) ) ( (0,-5) ) ( (2,3) ) ( (5,0) )

Explanation:

Step1: Understand the transformation rules

First, reverse the rotation. The rule for a \(90^{\circ}\) counter - clockwise rotation about the origin \(R_{0,90^{\circ}}(x,y)=(-y,x)\), so the reverse rotation \(R_{0, - 90^{\circ}}(x,y)=(y, - x)\). Second, reverse the translation. The translation rule \(T_{-1,1}(x,y)=(x - 1,y + 1)\), so the reverse translation \(T_{1,-1}(x,y)=(x + 1,y - 1)\).

Step2: Apply the reverse - transformation

Let the coordinates of \(B'\) (after translation but before rotation) be \((x',y')\). Since \(R_{0,90^{\circ}}(x',y')=(-y',x')\) and \(R_{0,90^{\circ}}(x')=-4,x' = 1\) (because \(B''(-4,1)\) is the result of rotation). Solving \(-y'=-4\) and \(x' = 1\) gives \(y' = 4,x' = 1\). Then, since \(T_{-1,1}(x,y)=(x - 1,y + 1)=(x',y')=(1,4)\). Using the reverse - translation \(x=x'+1,y=y'-1\).

Step3: Calculate the coordinates

Substitute \(x' = 1,y' = 4\) into the reverse - translation formula. \(x=1 + 1=2,y=4-1 = 3\)

Answer:

\((2,3)\)