Question

The Coordinates Of A (5,2) B (1,5)AB is dilated about the point (5,5) with a scale factor of 2 what is the length of the dilation of AB?

Answer 1

We are going to find the coordinates of the points A'B', which are given by the dilation of AB about the point (5,5) with a scale factor of 2.

For doing so, we will determine the distance between the points A and (5,5), and we will multiply it by 2 (the scale factor). This will be the distance between A' and (5,5). We will do the same between the point B and (5,5).

As the lines x=5 and y=5 are perpendicular, the angle BPA is right (where P is the point (5,5)), we can find the distance of the point A'B' using the Pythagorean Theorem.

Now, we find the distance between A and P:

`\begin{gathered} d(A,P)=\sqrt[]{(5-5)^2+(2-5)^2} \\ =\sqrt[]{0^2+(-3)^2} \\ =\sqrt[]{3^2} \\ =3 \end{gathered}`

And the distance between B and P:

`\begin{gathered} d(B,P)=\sqrt[]{(1-5)^2+(5-5)^2} \\ =\sqrt[]{(-4)^2+0^2} \\ =\sqrt[]{16} \\ =4 \end{gathered}`

Multiplying by the scale factor we obtain:

`\begin{gathered} d(A^(\prime),P)=2\cdot d(A,P)=2\cdot3=6 \\ d(B^(\prime),P)=2\cdot d(B,P)=2\cdot4=8 \end{gathered}`

And using the Pythagorean Theorem we obtain:

`\begin{gathered} d(A^(\prime),B^(\prime))=\sqrt[]{d(A^(\prime),P)^2+d(B^(\prime),P)^2} \\ =\sqrt[]{6^2+8^2} \\ =\sqrt[]{36+64} \\ =\sqrt[]{100} \\ =10 \end{gathered}`

This means that the length of the dilation of AB will be 10.