Q34. - Algebra 2

Question

LANDSCAPING

For exercises 33 and 34, use the following information.

A garden design is plotted on a coordinated grid. The original plan shows a fountain with vertices at $\left(-2,-2\right),\left(-6,-2\right),\left(-8,-5\right),\text{ and }\left(-4,-5\right)$. Changes to the plan now require that the fountain’s perimeter be the three-fourths that of the original.

34. The center of the fountain was at $\left(-5,-3.5\right)$. What will be the coordinate of the center after the changes in the plan have been made?

Step-by-Step Solution

Answer

The coordinate of the center after the changes in the plan is $\left(-\frac{15}{4},-\frac{21}{8}\right)$

Step 1 - Vertex matrix

An ordered pair $\left(x,y\right)$ is represented by a column matrix $\left[\begin{array}{c}x\\ y\\ \end{array}\right]$ which is known as vertex matrix of an ordered pair

For the coordinate of center as $\left(-5,-3.5\right)$, its vertex matrix is $\left[\begin{array}{c}-5\\ -3.5\\ \end{array}\right]=\left[\begin{array}{c}-5\\ -\frac{7}{2}\\ \end{array}\right]$

Step 2 - Vertex matrix after dilation

As perimeter is reduced to three-fourth of original quadrilateral, so vertex matrix of image is formed after dilation.

Vertex matrix after dilation is product of scale factor with given vertex matrix as shown below

$\frac{3}{4}\times \left[\begin{array}{c}-5\\ -\frac{7}{2}\\ \end{array}\right]=\left[\begin{array}{c}-\frac{15}{4}\\ -\frac{21}{8}\\ \end{array}\right]$

Step 3 - Coordinates of image after dilation

As, each coordinate is represented by a column of vertex matrix, so each column represents one coordinate.

Thus, coordinate of image after dilation is $\left(-\frac{15}{4},-\frac{21}{8}\right)$