Question

Find the scale factor of the line segment dilation. AB: endpoints (-6, -3) and (-3,-9) to A'B': endpoints at (-2, -1) and (-1, -3). A) -1/3 B) 1/3 C) 3D) -3

Answer 1

we know that the endpoints of AB are

`\begin{gathered} (x_1,y_1)=\mleft(-6,-3\mright) \\ (x_2,y_2)=(-3,-9) \end{gathered}`

and the distance formula is given by

`d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}`

By substituying these points, we have that

`d=\sqrt[]{(-3-(-6))^2+(-9-(-3))^2}`

which is equal to

`\begin{gathered} d=\sqrt[]{(-3+6)^2+(-9+3)^2} \\ d=\sqrt[]{3^2+(-6)^2} \end{gathered}`

then

`\begin{gathered} d=\sqrt[]{9+36} \\ d=\sqrt[]{45} \\ d=\sqrt[]{9\cdot5} \\ d=\sqrt[]{9}\cdot\sqrt[]{5} \\ d=3\sqrt[]{5}\ldots..(A) \end{gathered}`

On the other hand, if

`\begin{gathered} (x_1,y_1)=(-2,-1) \\ (x_2,y_2)=(-1,-3) \end{gathered}`

similarly to the previous case, the distance between the endpoint for A'B' is

`d=\sqrt[]{(-1-(-2))^2+(-3-(-1))^2}`

which is equal to

`\begin{gathered} d=\sqrt[]{(-1+2)^2+(-3+1)^2} \\ d=\sqrt[]{1^2+(-2)^2} \\ d=\sqrt[]{1+4} \\ d=\sqrt[]{5}\ldots..(B) \end{gathered}`

Now, by comparing equation A and equation B, we can see that, the scale factor is 1/3.

Then, the answer is B.