Figure 2 is the image of Figure 1 after rotation by an angle of 90degrees counter-clockwise
Step 1: Write out the coordinates of EFGH and the coordinates of KLMN,
[tex]\begin{gathered} \text{ The coordinates of E }=(1,2) \\ \text{ The coordinates of F }=(4,3) \\ \text{ The coordinates of G }=(3,1) \\ \text{ The coordinates of H }=(1,1) \end{gathered}[/tex][tex]\begin{gathered} \text{ The coordinates of K }=(-2,1) \\ \text{ The coordinates of L }=(-3,4) \\ \text{ The coordinates of M }=(-1,3) \\ \text{ The coordinates of N }=(-1,1) \end{gathered}[/tex]Step 2: Write out the formula for the rotation of a point (x,y) by 90° counter-clockwise
[tex]R(x,y)\to(-y,x)[/tex]Step 3: Show that the points KLMN are the images of R on EFGH
[tex]\begin{gathered} R\text{ on E }=R(1,2)=(-2,1)=\text{ the coordinates of K} \\ R\text{ on F }=R(4,3)=(-3,4)=\text{ the coordinates of L} \\ R\text{ on G }=R(3,1)=(-1,3)=\text{ the coordinates of M} \\ R\text{ on H }=R(1,1)=(-1,1)=\text{ the coordinates of N} \end{gathered}[/tex]Hence, Figure 2 is the image of Figure 1 on Rotation by 90 degrees counter-clockwise
Since rotation preserves the shape of a figure, then the measures of corresponding angles are equal.
Therefore,
m
The third Choice is correct