Chapter 14.5, Problem 27WE

To determine

To find: The coordinates of the center $\left(a,b\right)$ and the value of k .

Answer to Problem 27WE

The coordinates of the center $\left(a,b\right)$ are $\left(4,2\right)$ and the value of k is 3.

Explanation of Solution

Given information: The point $A\left(3,4\right)$ maps to $A^{\prime }\left(1,8\right)$ and $B\left(3,2\right)$ to $B^{\prime }\left(1,2\right)$ with scale factor k and dilation with center $\left(a,b\right)$ .

Calculation:

If transformation is related to similarity rather than congruence is known as dilation. The dilation $D_{O,k}$ has center $O$ and nonzero scale factor $k$ .

The dilation $D_{O,k}$ maps at any point $P$ to a point $P^{\prime }$ can be determined as follows:

(1) If $k$ is greater than 0, $P^{\prime }$ lie on $\overline{OP}$ and $O{P}^{\prime }=k\cdot OP$ .

(2) If $k$ is less than 0, $P\text{'}$ lies on the ray opposite to $\overline{OP}$ and $O{P}^{\prime }=\left|k\right|\cdot OP$ .

(3) The center $O$ is its own image is:

For $\left|k\right|>1$ , dilation is called an expansion and for $\left|k\right|<1$ , the dilation is called an contraction.

So, $A^{\prime }{B}^{\prime }=\left|k\right|\cdot AB$ .

The formula to find the distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is:

  $\text{distance}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The distance between point A and B is:

  $\begin{array}{c}AB=\sqrt{{\left(3-3\right)}^2+{\left(2-4\right)}^2}\\ =\sqrt{0+{\left(-2\right)}^2}\\ =\sqrt{4}\\ =2\end{array}$

The distance between point $A^{\prime }$ and $B^{\prime }$ is:

  $\begin{array}{c}{A}^{\prime }{B}^{\prime }=\sqrt{{\left(1-1\right)}^2+{\left(2-8\right)}^2}\\ =\sqrt{0+{\left(-6\right)}^2}\\ =\sqrt{36}\\ =6\end{array}$

Now, substitute the values in $A^{\prime }{B}^{\prime }=\left|k\right|\cdot AB$ .

  $\begin{array}{c}6=\left|k\right|\cdot 2\\ \left|k\right|=3\end{array}$

It is given that the center is $\left(a,b\right)$ . Thus, $O{A}^{\prime }=\left|k\right|\cdot OA$ .

The distance between point O and A is:

  $\begin{array}{c}OA=\sqrt{{\left(3-a\right)}^2+{\left(4-b\right)}^2}\\ =\sqrt{9+a^2-6a+16+b^2-8b}\\ =\sqrt{25+a^2+b^2-6a-8b}\end{array}$

The distance between point O and $A^{\prime }$ is:

  $\begin{array}{c}O{A}^{\prime }=\sqrt{{\left(1-a\right)}^2+{\left(8-b\right)}^2}\\ =\sqrt{1+a^2-2a+64+b^2-16b}\\ =\sqrt{65+a^2+b^2-2a-16b}\end{array}$

Now,

  $\begin{array}{c}O{A}^{\prime }=\left|k\right|\cdot OA\\ \sqrt{65+a^2+b^2-2a-16b}=3\left(\sqrt{25+a^2+b^2-6a-8b}\right)\\ 65+a^2+b^2-2a-16b=9\left(25+a^2+b^2-6a-8b\right)\\ 8a^2+8b^2+160-52a-56b=0\mathrm{......}\left(1\right)\end{array}$

The distance between point O and B is:

  $\begin{array}{c}OB=\sqrt{{\left(3-a\right)}^2+{\left(2-b\right)}^2}\\ =\sqrt{9+a^2-6a+4+b^2-4b}\\ =\sqrt{13+a^2+b^2-6a-4b}\end{array}$

The distance between point O and $B^{\prime }$ is:

  $\begin{array}{c}O{B}^{\prime }=\sqrt{{\left(1-a\right)}^2+{\left(2-b\right)}^2}\\ =\sqrt{1+a^2-2a+4+b^2-4b}\\ =\sqrt{5+a^2+b^2-2a-4b}\end{array}$

Now,

  $\begin{array}{c}O{B}^{\prime }=\left|k\right|\cdot OB\\ \sqrt{13+a^2+b^2-6a-4b}=3\left(\sqrt{5+a^2+b^2-2a-4b}\right)\\ 13+a^2+b^2-6a-4b=9\left(5+a^2+b^2-2a-4b\right)\\ 8a^2+8b^2+112-52a-32b=0\mathrm{......}\left(2\right)\end{array}$

Compare equation (1) and (2), then solve as:

  $\begin{array}{c}8a^2+8b^2+160-52a-56b=8a^2+8b^2+112-52a-32b\\ 160-56b=112-32b\\ 48=24b\\ b=2\end{array}$

Substitute 2 for b in the equation (2).

  $\begin{array}{c}8a^2+8{\left(2\right)}^2+112-52a-32\left(2\right)=0\\ 8a^2+32+112-52a-64=0\\ 8a^2-52a+80=0\\ 2a^2-13a+20=0\end{array}$

Simplify further.

  $\begin{array}{c}2a\left(a-4\right)-5\left(a-4\right)=0\\ \left(2a-5\right)\left(a-4\right)=0\\ a=4\end{array}$

Therefore, the coordinates of the center $\left(a,b\right)$ are $\left(4,2\right)$ and the value of k is 3.