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 $\boldsymbol{3}$ unit(s). the distance in the $y$-coordinates from $a(-2, 2)$ to the center of dilation $f(1, 1)$ is $\boldsymbol{\text{dropdown: -1, 0, 1, 2}}$ unit(s). the vertex $a$ of the image $\boldsymbol{\text{dropdown: (-8, 4) with cross mark on -1}}$.

Explanation:

Step1: Calculate y - coordinate distance

To find the distance in the y - coordinates between \(A(-2,2)\) and \(F(1,1)\), we use the formula for the distance between two points in the y - direction, which is \(|y_A - y_F|\). Substituting the values, we get \(|2 - 1|=1\).

Step2: Find the coordinates of \(A'\)

The dilation rule is \(D_{F,3}(x,y)\), which means we scale the distance from the center of dilation \(F(1,1)\) by a factor of 3.
First, for the x - coordinate: The distance from \(A\) to \(F\) in the x - direction is \(1-(-2) = 3\) (we can also use absolute value \(|-2 - 1| = 3\)). After dilation, the new distance from \(F\) in the x - direction is \(3\times3=9\)? Wait, no, the correct way is to use the formula for dilation about a point \((h,k)\): the image of a point \((x,y)\) under dilation with center \((h,k)\) and scale factor \(r\) is \((h + r(x - h),k + r(y - k))\).
So for point \(A(-2,2)\) and center \(F(1,1)\) with \(r = 3\):
The x - coordinate of \(A'\) is \(1+3\times(-2 - 1)=1+3\times(-3)=1 - 9=-8\)
The y - coordinate of \(A'\) is \(1+3\times(2 - 1)=1 + 3\times1=4\)
But first, the distance in y - coordinates: \(|2 - 1| = 1\).

Answer:

The distance in the y - coordinates from \(A(-2,2)\) to \(F(1,1)\) is \(\boldsymbol{1}\) unit(s).