Chapter 9.4, Problem 72AYU

a.

To determine

To find: the new vertices of the parallelogram $A^{\prime },B^{\prime },C^{\prime },D^{\prime }$ if it is translated by $v=\langle 3,-2\rangle$ .

Answer to Problem 72AYU

  $\left(0,-2\right),\left(2,-4\right),\left(6,-1\right),\left(4,1\right)$ .

Explanation of Solution

Given:

The field of computer graphics utilize vectors to compute translation of point.

The points $\left(-3,0\right)\left(-1,-2\right)\left(3,1\right)\left(1,3\right)$ are the vertices of a parallelogram ABCD.

Calculation:

Since the difference between two points is a vector, therefore sum of two point and a vector is a point.

  $\begin{array}{l}\left(-3,0\right)\to \left(-3,0\right)+\langle 3,-2\rangle =\left(0,-2\right)\\ \left(-1,-2\right)\to \left(-1,-2\right)+\langle 3,-2\rangle =\left(2,-4\right)\\ \left(3,1\right)\to \left(3,1\right)+\langle 3,-2\rangle =\left(6,-1\right)\\ \left(1,3\right)\to \left(1,3\right)+\langle 3,-2\rangle =\left(4,1\right)\end{array}$ .

Hence, the vertices of the parallelogram are $\left(0,-2\right),\left(2,-4\right),\left(6,-1\right),\left(4,1\right)$ .

b.

To determine

To find: the new vertices of the parallelogram $A^{\prime },B^{\prime },C^{\prime },D^{\prime }$ if it is translated by $-\frac{1}{2}v$ .

Answer to Problem 72AYU

  $\left(-4.5,1\right),\left(-2.5,-1\right),\left(1.5,2\right),\left(-0.5,4\right)$ .

Explanation of Solution

Given:

The field of computer graphics utilize vectors to compute translation of point.

The points $\left(-3,0\right)\left(-1,-2\right)\left(3,1\right)\left(1,3\right)$ are the vertices of a parallelogram ABCD.

Calculation:

Vector $\frac{1}{2}v$ has coordinates $\langle -1,5,1\rangle$ . Now add each vector to the each of the vertices of the parallelogram.

  $\begin{array}{l}\left(-3,0\right)\to \left(-3,0\right)+\langle -1,5,1\rangle =\left(-4.5,1\right)\\ \left(-1,-2\right)\to \left(-1,-2\right)+\langle -1,5,1\rangle =\left(-2.5,-1\right)\\ \left(3,1\right)\to \left(3,1\right)+\langle -1,5,1\rangle =\left(1.5,2\right)\\ \left(1,3\right)\to \left(1,3\right)+\langle -1,5,1\rangle =\left(-0.5,4\right)\end{array}$ .

Hence, the vertices of the parallelogram are $\left(-4.5,1\right),\left(-2.5,-1\right),\left(1.5,2\right),\left(-0.5,4\right)$ .