Question

what is the length of segment ac? triangle △abc is the image of △abc under a dilation with a scale factor of 2.

Explanation:

Step1: Use distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Assume the coordinates of point $A$ are $(x_A,y_A)$ and of point $C$ are $(x_C,y_C)$ from the grid. Count the horizontal and vertical distances. If we assume $A$ is at $(x_1,y_1)$ and $C$ is at $(x_2,y_2)$, by counting the grid - squares, the horizontal distance $\Delta x=x_2 - x_1$ and the vertical distance $\Delta y=y_2 - y_1$. Here, by counting the grid - squares, the horizontal distance between $A$ and $C$ is 6 units and the vertical distance is 2 units.

Step2: Apply the formula

$d=\sqrt{(6)^2+(2)^2}=\sqrt{36 + 4}=\sqrt{40}=2\sqrt{10}$ units. But we know that $\triangle A'B'C'$ is the image of $\triangle ABC$ under a dilation with a scale factor of 2. To find the length of the original segment $AC$, we use the fact that if $d'$ is the length of the dilated segment and $d$ is the length of the original segment and $k$ is the scale factor, $d'=k\times d$. Since we want to find $d$ and we know $d'$ (the length of $A'C'$ calculated above) and $k = 2$, we can also use the grid - counting in reverse. The length of $A'C'$ on the grid (for the dilated triangle) has a horizontal displacement of 6 and a vertical displacement of 2. For the original triangle, the horizontal displacement of $AC$ is 3 and the vertical displacement is 1.
$d=\sqrt{(3)^2+(1)^2}=\sqrt{9 + 1}=\sqrt{10}$ units.

Answer:

$\sqrt{10}$ units