Combined equation of a pair of lines

We already know that an equation of the form $ax + by + c = 0$ (with $a, b, c \in \mathbb{R}$ and $a, b$ not both zero) describes a single straight line in the $XY$-plane. This section asks what happens when we want a single equation to describe two lines at once.

Let $u \equiv a_1x + b_1y + c_1$ and $v \equiv a_2x + b_2y + c_2$, so that $u = 0$ and $v = 0$ are two given lines. An equation that represents both of these lines together is called the combined equation (also called the joint equation) of the pair. The claim is that the product $uv = 0$ is exactly this combined equation.

Theorem 4.1. The equation $uv = 0$ represents the combined equation of the lines $u = 0$ and $v = 0$.

Proof. We must show two things: every point on either line satisfies $uv = 0$, and every point satisfying $uv = 0$ lies on one of the two lines.

First, take any point $P(x_1, y_1)$ on the line $u = 0$, so $a_1x_1 + b_1y_1 + c_1 = 0$. Then the product $(a_1x_1+b_1y_1+c_1)(a_2x_1+b_2y_1+c_2)$ has a first factor equal to $0$, so the whole product is $0$ regardless of the second factor. Hence $P$ satisfies $uv = 0$. The same argument with the roles of $u$ and $v$ swapped shows every point of $v = 0$ also satisfies $uv = 0$.

Conversely, take any point $R(x', y')$ that satisfies $uv = 0$, i.e. $(a_1x'+b_1y'+c_1)(a_2x'+b_2y'+c_2) = 0$. A product of two real numbers is zero only if at least one factor is zero, so either $a_1x'+b_1y'+c_1 = 0$ (meaning $R$ lies on $u = 0$) or $a_2x'+b_2y'+c_2 = 0$ (meaning $R$ lies on $v = 0$).

Since the solution set of $uv = 0$ is exactly the union of the two lines, $uv = 0$ is their combined equation. $\blacksquare$

Remarks. (1) The combined equation of a pair of lines is also called the joint equation. (2) The individual equations $u = 0$ and $v = 0$ are called the separate equations of the two lines that make up $uv = 0$.

Worked Examples

Example 1. Find the combined equation of the lines $x + y - 2 = 0$ and $2x - y + 2 = 0$.

Multiply the two expressions: $(x+y-2)(2x-y+2) = 0$. Expanding term by term, $x(2x-y+2) + y(2x-y+2) - 2(2x-y+2) = 2x^2 - xy + 2x + 2xy - y^2 + 2y - 4x + 2y - 4$. Collecting like terms gives the combined equation $2x^2 + xy - y^2 - 2x + 4y - 4 = 0$.

Example 2. Find the combined equation of the lines $x - 2 = 0$ and $y + 2 = 0$.

$(x-2)(y+2) = 0$ expands to $xy + 2x - 2y - 4 = 0$, which is the required combined equation.

Example 3. Find the combined equation of the lines $x - 2y = 0$ and $x + y = 0$.

$(x-2y)(x+y) = 0$ expands to $x^2 + xy - 2xy - 2y^2 = x^2 - xy - 2y^2 = 0$, the required combined equation.

Example 4. Find the separate equations of the lines represented by $x^2 - y^2 + x + y = 0$.

Group the difference of squares and the linear terms: $x^2 - y^2 + x + y = (x+y)(x-y) + (x+y) = (x+y)\big[(x-y) + 1\big] = (x+y)(x-y+1)$. Setting this product to zero, the separate equations are $x + y = 0$ and $x - y + 1 = 0$.