Question

you do 1
triangle xyz has vertices x(0, 2), y(4, 4), and z(3, -1). find the image of △xyz after a rotation of 180° about (2, -3).

Explanation:

Step1: Use the rotation formula

The formula for rotating a point $(x,y)$ 180° about a center $(a,b)$ is $(2a - x,2b - y)$.

Step2: Find the image of point X

For $X(0,2)$ with $(a = 2,b=-3)$, we have $x_1=2\times2 - 0=4$ and $y_1=2\times(-3)-2=-6 - 2=-8$. So the image of $X$ is $X'(4,-8)$.

Step3: Find the image of point Y

For $Y(4,4)$ with $(a = 2,b = - 3)$, we have $x_2=2\times2-4 = 0$ and $y_2=2\times(-3)-4=-6 - 4=-10$. So the image of $Y$ is $Y'(0,-10)$.

Step4: Find the image of point Z

For $Z(3,-1)$ with $(a = 2,b=-3)$, we have $x_3=2\times2 - 3=1$ and $y_3=2\times(-3)-(-1)=-6 + 1=-5$. So the image of $Z$ is $Z'(1,-5)$.

Answer:

The vertices of the image of $\triangle XYZ$ are $X'(4,-8),Y'(0,-10),Z'(1,-5)$