Question

the dilation rule ( d_{f,3}(x, y) ) is applied to ( \triangle abc ), where the center of dilation is at ( f(1, 1) ). the distance in the x-coordinates from ( a(-2, 2) ) to the center of dilation ( f(1, 1) ) is (square) unit(s). the distance in the y-coordinates from ( a(-2, 2) ) to the center of dilation ( f(1, 1) ) is (square) unit(s). the vertex ( a ) of the image is (square) with options ((-8, 4)), ((-5, 3)), ((-3, 1)), ((2, 2)).

Explanation:

Part 1: Distance in x - coordinates

Step 1: Recall the formula for horizontal distance

The horizontal (x - coordinate) distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by \(|x_2 - x_1|\). Here, the coordinates of \(A\) are \((-2,2)\) and the coordinates of \(F\) are \((1,1)\). So we calculate \(|1-(-2)|\).
\(1-(-2)=1 + 2=3\)

Step 2: Determine the absolute value

Since the distance is non - negative, \(|3| = 3\).

Step 1: Recall the formula for vertical distance

The vertical (y - coordinate) distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by \(|y_2 - y_1|\). Here, the coordinates of \(A\) are \((-2,2)\) and the coordinates of \(F\) are \((1,1)\). So we calculate \(|1 - 2|\).
\(1-2=-1\)

Step 2: Determine the absolute value

Since the distance is non - negative, \(|-1|=1\).

Step 1: Recall the dilation rule

The dilation rule \(D_{F,3}(x,y)\) means that we first find the vector from the center of dilation \(F(1,1)\) to the point \(A(-2,2)\), then multiply that vector by the scale factor \(3\), and then add it back to the center of dilation \(F\).

The vector from \(F\) to \(A\) in the x - direction is \(x_A - x_F=-2 - 1=-3\), and in the y - direction is \(y_A - y_F=2 - 1 = 1\).

Step 2: Apply the scale factor

After multiplying by the scale factor \(3\), the x - component of the vector becomes \(3\times(-3)=-9\) and the y - component becomes \(3\times1 = 3\).

Step 3: Find the coordinates of \(A'\)

To find the coordinates of \(A'\), we add these scaled vectors to the coordinates of \(F\).

The x - coordinate of \(A'\) is \(x_F+(-9)=1+(-9)=-8\)

The y - coordinate of \(A'\) is \(y_F + 3=1 + 3=4\)

So the coordinates of \(A'\) are \((-8,4)\)

Answer:

3

Part 2: Distance in y - coordinates