Question

triangle xyz is the image after a dilation of △abc.

1. determine the scale factor.
2. determine the center of dilation.
3. explain how you could verify that the ratio of corresponding sides is constant.

Explanation:

Step1: Recall the formula for scale - factor

The scale factor $k$ for a dilation from a point $(x_1,y_1)$ to $(x_2,y_2)$ with center of dilation $(a,b)$ is given by $k=\frac{x_2 - a}{x_1 - a}=\frac{y_2 - b}{y_1 - b}$. First, assume the center of dilation is the origin $(0,0)$ (we will verify this later). For a point $(x,y)$ in the original figure and $(x',y')$ in the dilated figure, $k=\frac{x'}{x}=\frac{y'}{y}$. Let's take point $A(1,8)$ and its corresponding point $X(-6,4)$.

Step2: Calculate the scale - factor for the x - coordinates

$k_x=\frac{-6}{1}=-6$

Step3: Calculate the scale - factor for the y - coordinates

$k_y=\frac{4}{8}=\frac{1}{2}$
Since $k_x
eq k_y$, our assumption of the origin as the center of dilation is wrong. Let the center of dilation be $(a,b)$. Using the formula for two corresponding points $A(1,8)$ and $X(-6,4)$:
$\frac{-6 - a}{1 - a}=\frac{4 - b}{8 - b}$
Let's use another pair of corresponding points, say $B(5,-5)$ and $Y(-4,-2.5)$
$\frac{-4 - a}{5 - a}=\frac{-2.5 - b}{-5 - b}$
We can also use the property that if $(x_1,y_1)$ dilates to $(x_2,y_2)$ with center of dilation $(a,b)$ and scale factor $k$: $x_2 - a=k(x_1 - a)$ and $y_2 - b=k(y_1 - b)$
Let's assume the center of dilation is $(a,b)$. For point $A(1,8)$ and $X(-6,4)$:
$-6 - a=k(1 - a)$ and $4 - b=k(8 - b)$
For point $B(5,-5)$ and $Y(-4,-2.5)$:
$-4 - a=k(5 - a)$ and $-2.5 - b=k(-5 - b)$
Subtract the first equation of the $A - X$ pair from the first equation of the $B - Y$ pair:
$(-4 - a)-(-6 - a)=k((5 - a)-(1 - a))$
$2 = k\times4$, so $k = 0.5$

Step4: Find the center of dilation

Using the formula $x_2 - a=k(x_1 - a)$ with point $A(1,8)$ and $X(-6,4)$ and $k = 0.5$
$-6 - a=0.5(1 - a)$
$-6 - a=0.5-0.5a$
$-a + 0.5a=0.5 + 6$
$-0.5a=6.5$
$a=-13$
Using the formula $y_2 - b=k(y_1 - b)$ with $k = 0.5$, $y_1 = 8$ and $y_2 = 4$
$4 - b=0.5(8 - b)$
$4 - b=4-0.5b$
$-b + 0.5b=4 - 4$
$-0.5b = 0$
$b = 0$
The center of dilation is $(-13,0)$

1. # Answer:

The scale factor $k = 0.5$

1. # Answer:

The center of dilation is $(-13,0)$

1. # Answer:

To verify that the ratio of corresponding side lengths is constant, we can calculate the lengths of corresponding sides of the two triangles.
The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$
For $\triangle ABC$, the length of $AB$ with $A(1,8)$ and $B(5,-5)$:
$d_{AB}=\sqrt{(5 - 1)^2+(-5 - 8)^2}=\sqrt{16 + 169}=\sqrt{185}$
For $\triangle XYZ$, the length of $XY$ with $X(-6,4)$ and $Y(-4,-2.5)$:
$d_{XY}=\sqrt{(-4+6)^2+(-2.5 - 4)^2}=\sqrt{4+( - 6.5)^2}=\sqrt{4 + 42.25}=\sqrt{46.25}$
The ratio $\frac{d_{XY}}{d_{AB}}=\frac{\sqrt{46.25}}{\sqrt{185}}=\frac{1}{2}$
We can do this for other pairs of corresponding sides (e.g., $BC$ and $YZ$, $AC$ and $XZ$) and check that the ratio is $\frac{1}{2}$ each time.

Answer:

Step1: Recall the formula for scale - factor

The scale factor $k$ for a dilation from a point $(x_1,y_1)$ to $(x_2,y_2)$ with center of dilation $(a,b)$ is given by $k=\frac{x_2 - a}{x_1 - a}=\frac{y_2 - b}{y_1 - b}$. First, assume the center of dilation is the origin $(0,0)$ (we will verify this later). For a point $(x,y)$ in the original figure and $(x',y')$ in the dilated figure, $k=\frac{x'}{x}=\frac{y'}{y}$. Let's take point $A(1,8)$ and its corresponding point $X(-6,4)$.

Step2: Calculate the scale - factor for the x - coordinates

$k_x=\frac{-6}{1}=-6$

Step3: Calculate the scale - factor for the y - coordinates

$k_y=\frac{4}{8}=\frac{1}{2}$
Since $k_x
eq k_y$, our assumption of the origin as the center of dilation is wrong. Let the center of dilation be $(a,b)$. Using the formula for two corresponding points $A(1,8)$ and $X(-6,4)$:
$\frac{-6 - a}{1 - a}=\frac{4 - b}{8 - b}$
Let's use another pair of corresponding points, say $B(5,-5)$ and $Y(-4,-2.5)$
$\frac{-4 - a}{5 - a}=\frac{-2.5 - b}{-5 - b}$
We can also use the property that if $(x_1,y_1)$ dilates to $(x_2,y_2)$ with center of dilation $(a,b)$ and scale factor $k$: $x_2 - a=k(x_1 - a)$ and $y_2 - b=k(y_1 - b)$
Let's assume the center of dilation is $(a,b)$. For point $A(1,8)$ and $X(-6,4)$:
$-6 - a=k(1 - a)$ and $4 - b=k(8 - b)$
For point $B(5,-5)$ and $Y(-4,-2.5)$:
$-4 - a=k(5 - a)$ and $-2.5 - b=k(-5 - b)$
Subtract the first equation of the $A - X$ pair from the first equation of the $B - Y$ pair:
$(-4 - a)-(-6 - a)=k((5 - a)-(1 - a))$
$2 = k\times4$, so $k = 0.5$

Step4: Find the center of dilation

Using the formula $x_2 - a=k(x_1 - a)$ with point $A(1,8)$ and $X(-6,4)$ and $k = 0.5$
$-6 - a=0.5(1 - a)$
$-6 - a=0.5-0.5a$
$-a + 0.5a=0.5 + 6$
$-0.5a=6.5$
$a=-13$
Using the formula $y_2 - b=k(y_1 - b)$ with $k = 0.5$, $y_1 = 8$ and $y_2 = 4$
$4 - b=0.5(8 - b)$
$4 - b=4-0.5b$
$-b + 0.5b=4 - 4$
$-0.5b = 0$
$b = 0$
The center of dilation is $(-13,0)$

1. # Answer:

The scale factor $k = 0.5$

1. # Answer:

The center of dilation is $(-13,0)$

1. # Answer:

To verify that the ratio of corresponding side lengths is constant, we can calculate the lengths of corresponding sides of the two triangles.
The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$
For $\triangle ABC$, the length of $AB$ with $A(1,8)$ and $B(5,-5)$:
$d_{AB}=\sqrt{(5 - 1)^2+(-5 - 8)^2}=\sqrt{16 + 169}=\sqrt{185}$
For $\triangle XYZ$, the length of $XY$ with $X(-6,4)$ and $Y(-4,-2.5)$:
$d_{XY}=\sqrt{(-4+6)^2+(-2.5 - 4)^2}=\sqrt{4+( - 6.5)^2}=\sqrt{4 + 42.25}=\sqrt{46.25}$
The ratio $\frac{d_{XY}}{d_{AB}}=\frac{\sqrt{46.25}}{\sqrt{185}}=\frac{1}{2}$
We can do this for other pairs of corresponding sides (e.g., $BC$ and $YZ$, $AC$ and $XZ$) and check that the ratio is $\frac{1}{2}$ each time.