Conditions for Homogeneity

Line
The line $$L = \alpha_1 x_1 + \alpha_2 x_2 = \beta$$ is homogeneous iff $$\beta = 0$$.

Plane
The plane $$P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$$ is homogeneous iff $$\gamma = 0$$.

Proof for Line

 * Let the line $$L = \alpha_1 x_1 + \alpha_2 x_2 = \beta$$ be homogeneous.

Then the origin $$\left({0, 0}\right)$$ lies on the line $$L$$.

That is, $$\alpha_1 0 + \alpha_2 0 = \beta \Longrightarrow \beta = 0$$.


 * Let the equation of $$L$$ be $$L = \alpha_1 x_1 + \alpha_2 x_2 = 0$$.

Then $$0 = \alpha_1 0 + \alpha_2 0 \in L$$ and so $$\left({0, 0}\right)$$ lies on the line $$L$$.

Hence $$L$$ is homogeneous.

Proof for Plane

 * Let the plane $$P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$$ be homogeneous.

Then the origin $$\left({0, 0, 0}\right)$$ lies on the plane $$P$$.

That is, $$\alpha_1 0 + \alpha_2 0 + \alpha_3 0= \gamma \Longrightarrow \gamma = 0$$.


 * Let the equation of $$P$$ be $$P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = 0$$.

Then $$0 = \alpha_1 0 + \alpha_2 0 + \alpha_3 0 \in P$$ and so $$\left({0, 0, 0}\right)$$ lies on the plane $$P$$.

Hence $$P$$ is homogeneous.