Question

you do 3 triangle fgh has vertices f(-3, 4), g(2, 0), and h(-1, -2). find the image of △fgh after a rotation of 180° about (-3, -6).

Explanation:

Step1: Recall rotation formula

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

Step2: Find image of point F

For point $F(-3,4)$ with center of rotation $(-3,-6)$, we substitute $x=-3,y = 4,a=-3,b=-6$ into the formula:
$2a - x=2\times(-3)-(-3)=-6 + 3=-3$
$2b - y=2\times(-6)-4=-12 - 4=-16$
So $F'(-3,-16)$.

Step3: Find image of point G

For point $G(2,0)$ with center of rotation $(-3,-6)$, we substitute $x = 2,y = 0,a=-3,b=-6$ into the formula:
$2a - x=2\times(-3)-2=-6-2=-8$
$2b - y=2\times(-6)-0=-12$
So $G'(-8,-12)$.

Step4: Find image of point H

For point $H(-1,-2)$ with center of rotation $(-3,-6)$, we substitute $x=-1,y=-2,a=-3,b=-6$ into the formula:
$2a - x=2\times(-3)-(-1)=-6 + 1=-5$
$2b - y=2\times(-6)-(-2)=-12 + 2=-10$
So $H'(-5,-10)$.

Answer:

The vertices of the image of $\triangle FGH$ are $F'(-3,-16),G'(-8,-12),H'(-5,-10)$