Question

which of the following reflective symmetries apply to the nonagon?
symmetry
applies to the figure?
reflective symmetry over the line ( y = -x )
reflection symmetry over the line ( y = -\frac{2}{3}x )

Explanation:

Step1: Understand reflective symmetry

Reflective symmetry means that when a figure is reflected over a line, the reflected image coincides with the original figure.

Step2: Analyze reflection over \(y = -x\)

For a non - regular nonagon (assuming the given nonagon is symmetric in a way that \(y=-x\) is a line of symmetry). Visually, if we consider the transformation of points \((x,y)\) to \((-y, -x)\) (reflection over \(y=-x\)), and check if the nonagon maps onto itself. From the graph, we can assume that the nonagon is constructed in a way that this reflection works.

Step3: Analyze reflection over \(y=-\frac{2}{3}x\)

For a general point \((x,y)\) reflected over the line \(y = mx + c\) (here \(c = 0,m=-\frac{2}{3}\)), the formula for reflection is more complex. But geometrically, if we pick a few non - vertex points on the nonagon (say a point \((3, - 2)\) which lies on \(y =-\frac{2}{3}x\)) and check the surrounding points. A nonagon (even if symmetric in some ways) is not symmetric about a non - axis - aligned line with a non - unit slope like \(y =-\frac{2}{3}x\) in a standard grid - based construction (assuming the nonagon is constructed with vertices at integer or half - integer coordinates in a simple geometric way).

Answer:

Reflective symmetry over the line \(y=-x\): Yes; Reflective symmetry over the line \(y =-\frac{2}{3}x\): No