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 dropdown unit(s). the distance in the ( y )-coordinates from ( a(-2, 2) ) to the center of dilation ( f(1, 1) ) is dropdown unit(s). the vertex ( a ) of the image dropdown with -1, 0, 1, 2 (partial text from dropdown visible).

Explanation:

Step1: Calculate x - coordinate distance

To find the distance in the x - coordinates between \(A(-2,2)\) and \(F(1,1)\), we use the formula for the distance between two x - coordinates: \(|x_2 - x_1|\). Here, \(x_1=-2\) (x - coordinate of \(A\)) and \(x_2 = 1\) (x - coordinate of \(F\)). So, \(|1-(-2)|=|1 + 2|=3\).

Step2: 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 y - coordinates: \(|y_2 - y_1|\). Here, \(y_1 = 2\) (y - coordinate of \(A\)) and \(y_2=1\) (y - coordinate of \(F\)). So, \(|1 - 2|=|-1| = 1\).

Step3: 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 \(F\) by a factor of 3.
For the x - coordinate of \(A'\): The distance from \(F\)'s x - coordinate (1) to \(A\)'s x - coordinate (-2) is 3 units (from Step 1). After dilation by a factor of 3, the new distance from \(F\)'s x - coordinate is \(3\times3 = 9\)? Wait, no. Wait, the formula for dilation with center \(F(h,k)\) and scale factor \(r\) is \((x',y')=(h + r(x - h),k + r(y - k))\).
So for point \(A(-2,2)\) and center \(F(1,1)\) and \(r = 3\):
\(x'=1+3\times(-2 - 1)=1+3\times(-3)=1-9=-8\)? Wait, maybe I made a mistake earlier. Wait, the distance in x - coordinates from \(A\) to \(F\) is \(1-(-2)=3\) (since we move from \(x=-2\) to \(x = 1\), the difference is \(1-(-2)=3\)). When we dilate with scale factor 3, the new x - coordinate of \(A'\) relative to \(F\) is \(3\times3=9\)? No, the formula for dilation about a center \((h,k)\) is \((x',y')=(h + r(x - h),k + r(y - k))\). So \(x=-2\), \(h = 1\), \(r = 3\). So \(x'=1+3\times(-2 - 1)=1-9=-8\), \(y'=1+3\times(2 - 1)=1 + 3\times1=4\). Wait, but maybe the question has a typo or I misread. Wait, the original problem's dropdown for \(A'\) has ( - 8,4)? Wait, the user's image shows a dropdown with - 1,0,1,2 and (3,4)? Wait, maybe I misread the coordinates. Wait, let's re - check the coordinates of \(A\). From the graph, \(A\) is at \((-2,2)\), \(F\) is at \((1,1)\).
The vector from \(F\) to \(A\) is \((-2 - 1,2 - 1)=(-3,1)\). When we dilate with scale factor 3, the vector from \(F\) to \(A'\) is \(3\times(-3,1)=(-9,3)\). Then \(A'\) has coordinates \((1-9,1 + 3)=(-8,4)\). But the dropdown in the user's image has - 1,0,1,2 and (3,4)? Maybe there is a mistake in the problem or my understanding. Wait, maybe the scale factor is 3, but the distance in x - coordinates from \(A\) to \(F\) is \(| - 2-1|=3\)? Wait, no, the x - coordinate of \(A\) is - 2, x - coordinate of \(F\) is 1. The difference is \(1-(-2)=3\). For the y - coordinate, \(2 - 1 = 1\). Then, when dilating with center \(F\) and scale factor 3, the new x - coordinate of \(A'\) is \(1+3\times(-2 - 1)=1-9=-8\), y - coordinate is \(1+3\times(2 - 1)=4\). But maybe the problem's dropdown is wrong, or I misread the coordinates. Wait, maybe \(A\) is at \((-1,2)\)? No, the graph shows \(A\) at \(x=-2\). Anyway, according to the calculation:

Answer:

The distance in the x - coordinates from \(A(-2,2)\) to \(F(1,1)\) is \(\boldsymbol{3}\) unit(s).
The distance in the y - coordinates from \(A(-2,2)\) to \(F(1,1)\) is \(\boldsymbol{1}\) unit(s).
For the vertex \(A'\), using the dilation formula \(D_{F,3}(x,y)=(1 + 3(x - 1),1+3(y - 1))\) (wait, no, the correct formula is \(D_{F,r}(x,y)=(h + r(x - h),k + r(y - k))\) where \(F(h,k)=(1,1)\) and \(r = 3\), \(x=-2\), \(y = 2\). So \(x'=1+3\times(-2 - 1)=1-9=-8\), \(y'=1+3\times(2 - 1)=4\). So \(A'\) is \((-8,4)\). But if we consider the distance in x - coordinates from \(A\) to \(F\) is 3, after dilation by 3, the new distance from \(F\) in x - direction is \(3\times3 = 9\), so \(x\) - coordinate of \(A'\) is \(1-9=-8\) (since \(A\) is to the left of \(F\)), and y - coordinate: distance from \(A\) to \(F\) in y - direction is 1, after dilation by 3, new distance is \(3\times1 = 3\), so \(y\) - coordinate of \(A'\) is \(1 + 3=4\) (since \(A\) is above \(F\)). So the coordinates of \(A'\) are \((-8,4)\). But the dropdown in the user's image has - 1,0,1,2 and (3,4), maybe there is a mistake in the problem's given options or my misinterpretation. However, based on the calculation:

The distance in x - coordinates: \(\boldsymbol{3}\)
The distance in y - coordinates: \(\boldsymbol{1}\)
The vertex \(A'\): \(\boldsymbol{(-8,4)}\) (but if we follow the wrong dropdown, maybe there is a mistake. But according to the dilation formula, this is the correct result.)