Question

In the diagram below, line segment AB has endpoints at A(-2,-6) and B( 3,-1) .Draw A'B' the image of AB after a counterclockwise rotation of 90° about the origin. Give the coordinates of its endpoints below. Is A'B' congruent to AB? Explain.

Answer 1

By definition, you know that when a point is rotated 90 degrees counterclockwise about the origin, point A (x, y) becomes A'(- y, x).

Then, if you rotate 90 degrees counterclockwise about the origin the points A and B you get:

$\begin{gathered} A(-2,6)\rightarrow A^{\prime}(-6,-2) \\ B(3,-1)\rightarrow B^{\prime}(-(-1),3)=B^{\prime}(1,3) \end{gathered}$

Now graphing points A and B and their respective rotations you have

Now, to know if the segments AB and A'B 'are congruent, you can find a measure of each one of them using the distance formula, which is

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

So, for the measure of segments AB you have

$\begin{gathered} (x_1,y_1)=(-2,-6) \\ (x_2,y_2)=(3,-1) \\ d_{AB}=\sqrt[]{(3-(-2))^2+(-1-(-6))^2} \\ d_{AB}=\sqrt[]{(3+2)^2+(-1+6)^2} \\ d_{AB}=\sqrt[]{(5)^2+(5)^2} \\ d_{AB}=\sqrt[]{25+25} \\ d_{AB}=\sqrt[]{50} \\ d_{AB}=7.07 \end{gathered}$

For the measure of segments A'B' you have

$\begin{gathered} (x_1,y_1)=(-6,-2) \\ (x_2,y_2)=(1,3) \\ d_{A^{\prime}B^{\prime}}=\sqrt[]{(1-(-6))^2+(3-(-2))^2} \\ d_{A^{\prime}B^{\prime}}=\sqrt[]{(1+6)^2+(3+2)^2} \\ d_{A^{\prime}B^{\prime}}=\sqrt[]{(7)^2+(5)^2} \\ d_{A^{\prime}B^{\prime}}=\sqrt[]{49+25} \\ d_{A^{\prime}B^{\prime}}=\sqrt[]{74} \\ d_{A^{\prime}B^{\prime}}=8.6 \end{gathered}$

Then, as you can see, the segments AB and A'B'do not have the same measure and therefore are not congruent.