Chapter 9.4, Problem 40HP

To determine

To find: The final coordinates of $P^{‴}{Q}^{‴}{R}^{″}{S}^{‴}$ .

Answer to Problem 40HP

The coordinates are $P^{‴}\left(1,-2\right)Q^{″}\left(2,1\right)R^{″}\left(-1,3\right)S^{‴}\left(-2,0\right)$ .

Explanation of Solution

Given:

The polygon PQRS is translated along the point $\left(3,-2\right)$ and is reflected in $y=-1$ and rotated $90°$ about the origin.

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

From the figure the coordinates of P, Q, R, S are $\left(-5,1\right),\left(-2,2\right),\left(0,-1\right),\left(-3,-2\right)$ .

The first step is to translate the figure PQRS along $\left(3,-2\right)$ as,

  $\begin{array}{l}\left(x,y\right)\to \left(x+3,y-2\right)\\ P\left(-5,1\right)\to P^{\prime }\left(-2,-1\right)\\ Q\left(-2,2\right)\to Q^{\prime }\left(1,0\right)\\ R\left(0,-1\right)\to R^{\prime }\left(3,-3\right)\end{array}$

Also,

  $S\left(-3,-2\right)\to S^{\prime }\left(0,-4\right)$

Now, reflect over the line $y=-1$ as,

  $\begin{array}{l}{P}^{\prime }\left(-2,-1\right)\to P^{″}\left(-2,-1\right)\\ Q^{\prime }\left(1,0\right)\to Q^{″}\left(1,-2\right)\\ R^{\prime }\left(3,-3\right)\to R^{″}\left(3,1\right)\\ S^{\prime }\left(0,-4\right)\to S^{″}\left(0,2\right)\end{array}$

Rotate $90°$ about the origin as shown below,

  $\begin{array}{l}{P}^{″}\left(-2,-1\right)\to P^{‴}\left(1,-2\right)\\ Q^{″}\left(1,-2\right)\to Q^{‴}\left(2,1\right)\\ R^{″}\left(3,1\right)\to Q^{‴}\left(-1,3\right)\\ S^{″}\left(0,2\right)\to S^{‴}\left(-2,0\right)\end{array}$

Thus, the coordinates are $P^{‴}\left(1,-2\right)Q^{″}\left(2,1\right)R^{″}\left(-1,3\right)S^{‴}\left(-2,0\right)$ .