Question

If each of (−2, 2), (0, 0) and (2, −2) is a solution of a linear equation in x and y, the equation is
(a) x − y = 0
(b) x + y = 0
(c) −2x + y = 0
(d) −x + 2y = 0

Answer 1

(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 $x+y=0$.