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 $\square$.

Explanation:

Step1: Find x - distance

The x - coordinate of \(A\) is \(-2\) and of \(F\) is \(1\). The distance is \(|1 - (-2)|=|1 + 2| = 3\).

Step2: Find y - distance

The y - coordinate of \(A\) is \(2\) and of \(F\) is \(1\). The distance is \(|2 - 1|=1\).

Step3: Find \(A'\)

For dilation with center \(F(1,1)\) and scale factor \(3\), the formula for a point \((x,y)\) is \((x',y')=(F_x+3(x - F_x),F_y + 3(y - F_y))\). For \(A(-2,2)\), \(x'=1+3(-2 - 1)=1+3\times(-3)=1 - 9=-8\), \(y'=1+3(2 - 1)=1 + 3\times1=4\). So \(A'=(-8,4)\).

Answer:

s:

• The distance in the x - coordinates: \(3\)
• The distance in the y - coordinates: \(1\)
• The vertex \(A'\): \((-8,4)\)