Each of the points (−2, 2), (0, 0), (2, −2) satisfies the linear equation

(b) $x+y=0$

Given points: $(-2,2),(0,0)$ and $(2,-2)$

We have to check which equation satisfies the given points.

Let us check for $(a) x-y=0$

Substituting $(-2,2)$ in the equation, we get: $x-y=-2-2=-4$

Substituting $(0,0)$ in the equation, we get: $x-y=0-0=0$

Substituting $(2,-2)$ in the equation, we get: $x-y=2+2=4$

So, the given points do not satisfy the equation.

Now, let us check (b) $x+y=0$

Substituting $(-2,2)$ in the equation, we get: $x+y=-2+2=0$

Substituting $(0,0)$ in the equation, we get: $x+y=0+0=0$

Substituting $(2,-2)$ in the equation, we get: $x+y=2-2=0$

So, the given points satisfy the equation.

Similarly, we can check other equations and find that the given points will not satisfy the equations.

Hence, each of $(-2,2),(0,0)$ and $(2,-2)$ is a solution of the linear equation $x+y=0$.