Chapter 4.4, Problem 16PPS

(a)

To determine

To find: The coordinates of $B^{\prime }$ .

Answer to Problem 16PPS

The coordinates of the vertices $B^{\prime }$ are $\left(6,-20\right)$ .

Explanation of Solution

Given:

The given table is shown in Table 1

Table 1

Quadrilateral $ABCD$	Quadrilateral $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$
$A\left(4,8\right)$	$A^{\prime }\left(-1,-2\right)$
$B\left(11,-10\right)$	$B^{\prime }\left(?,?\right)$
$C\left(?,?\right)$	$C^{\prime }\left(-17,-14\right)$
$D\left(-4,6\right)$	$D^{\prime }\left(-9,-4\right)$

Calculation:

Consider the translational matrix that shift $A\left(4,8\right)$ to $A^{\prime }\left(-1,-2\right)$ is,

  $\begin{array}{c}\left[\begin{array}{c}4\\ 8\end{array}\right]+\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-1\\ -2\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-1\\ -2\end{array}\right]-\left[\begin{array}{c}4\\ 8\end{array}\right]\\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-5\\ -10\end{array}\right]\end{array}$

This shows that the image $ABC$ is translated by adding -5 to the x coordinate and adding -10 to th3e y coordinate.

Then, the coordinate of vertices of the image is,

  $\left[\begin{array}{cccc}4& 11& a& -4\\ 8& -10& b& 6\end{array}\right]+\left[\begin{array}{cccc}-5& -5& -5& -5\\ -10& -10& -10& -10\end{array}\right]=\left[\begin{array}{cccc}-1& 6& -17& -9\\ -2& -20& -14& -4\end{array}\right]$

Then, the coordinates of the vertices $B^{\prime }$ are $\left(6,-20\right)$ .

(b)

To determine

To find: The coordinates of $C$ .

Answer to Problem 16PPS

The coordinates are $\left[-12,-4\right]$ .

Explanation of Solution

The matrix for the vertices of the image is,

  $\left[\begin{array}{cccc}4& 11& a& -4\\ 8& -10& b& 6\end{array}\right]+\left[\begin{array}{cccc}-5& -5& -5& -5\\ -10& -10& -10& -10\end{array}\right]=\left[\begin{array}{cccc}-1& 6& -17& -9\\ -2& -20& -14& -4\end{array}\right]$

Then, the coordinate of the vertices of $C$ is,

  $\begin{array}{l}\left[\begin{array}{c}a\\ b\end{array}\right]+\left[\begin{array}{c}-5\\ -10\end{array}\right]=\left[\begin{array}{c}-17\\ -14\end{array}\right]\\ \left[\begin{array}{c}a\\ b\end{array}\right]=\left[\begin{array}{c}-17\\ -14\end{array}\right]-\left[\begin{array}{c}-5\\ -10\end{array}\right]\\ \left[\begin{array}{c}a\\ b\end{array}\right]=\left[\begin{array}{c}-17+5\\ -14+10\end{array}\right]\\ \left[\begin{array}{c}a\\ b\end{array}\right]=\left[\begin{array}{c}-12\\ -4\end{array}\right]\end{array}$

Thus, the coordinatesare $\left[-12,-4\right]$ .