Conditions for Homogeneity

Theorem

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

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, $\alpha_1 0 + \alpha_2 0 = \beta \implies \beta = 0$.

• 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.

$\blacksquare$

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, $\alpha_1 0 + \alpha_2 0 + \alpha_3 0= \gamma \implies \gamma = 0$.

• 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.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 28$