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triangle xyz with vertices x(3, -2), y(6, 1), and z(5, -7): 180°. squar…

Question

triangle xyz with vertices x(3, -2), y(6, 1), and z(5, -7): 180°. square abcd with vertices a(-7, 5), b(-4, 7), c(-2, 4), and d(-5, 2): 90° counterclockwise

Explanation:

Step1: Recall rotation rules

For a 180 - degree rotation about the origin, the rule is $(x,y)\to(-x,-y)$. For a 90 - degree counter - clockwise rotation about the origin, the rule is $(x,y)\to(-y,x)$.

Step2: Rotate triangle XYZ

Given $X(3, - 2)$, $Y(6,1)$, $Z(5,-7)$. After a 180 - degree rotation, $X'(-3,2)$, $Y'(-6,-1)$, $Z'(-5,7)$.

Step3: Rotate square ABCD

Given $A(-7,5)$, $B(-4,7)$, $C(-2,4)$, $D(-5,2)$. After a 90 - degree counter - clockwise rotation, $A'(-5,-7)$, $B'(-7,-4)$, $C'(-4,-2)$, $D'(-2,-5)$.

Answer:

Vertices of rotated triangle $X'(-3,2)$, $Y'(-6,-1)$, $Z'(-5,7)$; vertices of rotated square $A'(-5,-7)$, $B'(-7,-4)$, $C'(-4,-2)$, $D'(-2,-5)$