Question

in the graph below, rectangle ( stuv ) is the image of ( stuv ) after a dilation.
what are the scale factor and center of the dilation?

Explanation:

Step1: Find the length of corresponding sides

For rectangle \(STUV\), the length of \(ST\) (horizontal side). The \(x -\)coordinate of \(S(-3,1)\) and \(T(5,1)\). Using the distance formula for horizontal line \(d=\vert x_2 - x_1\vert\), \(ST=\vert5-(-3)\vert = 8\).
For rectangle \(S'T'U'V'\), the length of \(S'T'\). The \(x -\)coordinate of \(S'(-9,5)\) and \(T'(3,5)\). Using the distance formula for horizontal line \(d=\vert x_2 - x_1\vert\), \(S'T'=\vert3-(-9)\vert = 12\).
The scale factor \(k\) of dilation is given by \(k=\frac{\text{length of image side}}{\text{length of pre - image side}}\). So \(k = \frac{S'T'}{ST}=\frac{12}{8}=\frac{3}{2}\).

Step2: Find the center of dilation

Let the center of dilation be \((a,b)\). Using the formula for dilation \((x',y')=(k(x - a)+a,k(y - b)+b)\).
Take a point \(S(-3,1)\) and its image \(S'(-9,5)\).
For the \(x -\)coordinate: \(-9=\frac{3}{2}(-3 - a)+a\).

$$
LATEXBLOCK0
$$

For the \(y -\)coordinate: \(5=\frac{3}{2}(1 - b)+b\).

$$
LATEXBLOCK1
$$

Another way:
We can also use the property that the center of dilation lies on the line joining a point and its image.
The slope of the line joining \(S(-3,1)\) and \(S'(-9,5)\) is \(m=\frac{5 - 1}{-9+3}=\frac{4}{-6}=-\frac{2}{3}\).
The equation of the line using point - slope form \(y - 1=-\frac{2}{3}(x + 3)\) or \(y=-\frac{2}{3}x - 1\).
The slope of the line joining \(V(-3,-3)\) and \(V'(-9,-1)\) is \(m=\frac{-1+3}{-9 + 3}=\frac{2}{-6}=-\frac{2}{3}\). The equation of the line using point - slope form \(y+3=-\frac{2}{3}(x + 3)\) or \(y=-\frac{2}{3}x-5\).
Let the center of dilation \((x,y)\) satisfy the proportion \(\frac{VS'}{VS}=\frac{3}{2}\).
The vector from \(V(-3,-3)\) to \(S(-3,1)\) is \(\overrightarrow{VS}=(0,4)\). The vector from \(V(-3,-3)\) to \(S'(-9,5)\) is \(\overrightarrow{VS'}=(-6,8)\).
If we consider the general formula for dilation. Let the center of dilation \((h,k)\).
For a point \(P(x,y)\) and its image \(P'(x',y')\), \(\frac{x'-h}{x - h}=\frac{y'-k}{y - k}=k\) (scale factor).
Take \(T(5,1)\) and \(T'(3,5)\)
\(\frac{3 - h}{5 - h}=\frac{3}{2}\)

$$
LATEXBLOCK2
$$

\(\frac{5 - k}{1 - k}=\frac{3}{2}\)

$$
LATEXBLOCK3
$$

Answer:

The scale factor is \(\frac{3}{2}\) and the center of dilation is \((9,-7)\)