Question

The endpoints of AB are A(9,4) and B(5,-4). The endpoints of its image after a dilation are A'(6,3) and B'(3,-3). Find the scale factor and explain each of your steps. This is for Geometry A. Please help!

Answer 1

Answer:

The scale factor is equal to $\frac{4}{3}$.

Step-by-step explanation:

The formula to calculate vector length if the following, where $| |$ denotes the length of a vector:

$|\stackrel{\rightarrow}{AB}| = \alpha |\stackrel{\rightarrow}{A'B'}|$

$\stackrel{\rightarrow}{AB} = \left(\begin{array}{c}9 - 5&\\4 - (-4)&\end{array}\right) = \left(\begin{array}{c}4&\\8&\end{array}\right)\\ \\ |\stackrel{\rightarrow}{AB}| = \sqrt{4^2 + 8^2} = \sqrt{16+64} = \sqrt{80}$

$\stackrel{\rightarrow}{A'B'} = \left(\begin{array}{c}6 - 3&\\3 - (-3)&\end{array}\right) = \left(\begin{array}{c}3&\\6&\end{array}\right)\\ \\ |\stackrel{\rightarrow}{AB}| = \sqrt{3^2 + 6^2} = \sqrt{9+36} = \sqrt{45}$

The scale factor formula is then defined by

$|\stackrel{\rightarrow}{AB}| = \alpha |\stackrel{\rightarrow}{A'B'}|\\ \\ \alpha = \frac{|\stackrel{\rightarrow}{AB}|}{|\stackrel{\rightarrow}{A'B'}|}\\ \\ \alpha = \frac{\sqrt{80}}{\sqrt{45}} = \sqrt{\frac{80}{45}} = \sqrt{\frac{16}{9}} = \frac{4}{3}$