Question

3 △abc is graphed on the coordinate plane. what is the length of the image of (overline{ab}) after a dilation of scale factor 3 centered at the origin? enter a number like 7 in the box.

Explanation:

Step1: Find the length of $\overline{AB}$ before dilation

Count the horizontal and vertical distances between points $A$ and $B$. Horizontally, the change in $x$ - values is from $x = 6$ to $x = 4$, a change of $2$ units. Vertically, the change in $y$ - values is from $y = 4$ to $y = 2$, a change of $2$ units. Using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, for points $A(6,4)$ and $B(4,2)$, we have $d_{AB}=\sqrt{(6 - 4)^2+(4 - 2)^2}=\sqrt{2^2+2^2}=\sqrt{4 + 4}=\sqrt{8}=2\sqrt{2}$.

Step2: Apply the dilation rule

When a figure is dilated by a scale factor $k$ centered at the origin, the length of any line - segment is multiplied by the scale factor. Here, $k = 3$. So the length of the dilated line - segment $d'_{AB}=k\times d_{AB}$. Substituting $k = 3$ and $d_{AB}=2\sqrt{2}$, we get $d'_{AB}=3\times2\sqrt{2}=6\sqrt{2}\approx 8.49$. But if we consider the Pythagorean relationship in a non - simplified way for the dilated right - triangle formed. The horizontal distance between the dilated points will be $3\times2 = 6$ and the vertical distance will be $3\times2=6$. Then using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, we have $d=\sqrt{6^2+6^2}=\sqrt{36 + 36}=\sqrt{72}=6\sqrt{2}\approx8.49$. Rounding to the nearest whole number, we can also calculate it as follows:
The right - triangle with sides adjacent to the right - angle for $\overline{AB}$ has side lengths $a = 2$ and $b = 2$. After dilation, the side lengths of the corresponding right - triangle are $a'=3\times2 = 6$ and $b'=3\times2 = 6$. By the Pythagorean theorem $c=\sqrt{a'^2 + b'^2}=\sqrt{6^2+6^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}\approx 8.49\approx8.5\approx9$.

Answer:

$6\sqrt{2}$ (or approximately $9$ if a whole - number answer is required)