Question

3.3b hw honors
rotate the figure 90° clockwise around (-1, -4). p(0, -6) q(5, -4) r(4, -7)

Explanation:

Step1: Translate points

First, translate the rotation - center and the points of the figure so that the rotation - center is at the origin. Let the rotation - center be $C(-1,-4)$. For a point $P(x,y)$, the translated point $P'(x',y')$ is given by $x'=x + 1$ and $y'=y + 4$.
For $P(0,-6)$: $x'_P=0 + 1=1$, $y'_P=-6 + 4=-2$.
For $Q(5,-4)$: $x'_Q=5 + 1=6$, $y'_Q=-4 + 4=0$.
For $R(4,-7)$: $x'_R=4 + 1=5$, $y'_R=-7 + 4=-3$.

Step2: Apply 90 - degree clockwise rotation formula

The formula for a 90 - degree clockwise rotation of a point $(x,y)$ about the origin is $(x,y)\to(y,-x)$.
For $P'(1,-2)$: After rotation, $P''(-2,-1)$.
For $Q'(6,0)$: After rotation, $Q''(0,-6)$.
For $R'(5,-3)$: After rotation, $R''(-3,-5)$.

Step3: Translate back

To get the final points, reverse the first - step translation. Let the final points be $P_f(x_f,y_f)$, $Q_f(x_f,y_f)$, $R_f(x_f,y_f)$. The reverse - translation formula is $x_f=x''-1$ and $y_f=y''-4$.
For $P''(-2,-1)$: $x_{P_f}=-2-1=-3$, $y_{P_f}=-1-4=-5$.
For $Q''(0,-6)$: $x_{Q_f}=0-1=-1$, $y_{Q_f}=-6-4=-10$.
For $R''(-3,-5)$: $x_{R_f}=-3-1=-4$, $y_{R_f}=-5-4=-9$.

Answer:

$P(-3,-5)$, $Q(-1,-10)$, $R(-4,-9)$