Every line in the Cartesian plane can be expressed in the general form Ax + By + C = 0. Suppose that the lines g and h have equations A_{1}x + B_{1}y + C_{1} = 0 and A_{2}x + B_{2}y + C_{2} = 0, respectively.

g ≡ A_{1}x + B_{1}y + C_{1} = 0 or g = 0

h ≡ A_{2}x + B_{2}y + C_{2} = 0 or h = 0

Let . It can be easily proved that g + λh = 0 is a line equation.

By substitution, we get the following:

g + λh = (A_{1} + λA_{2})x + (B_{1} + λB_{2})y + (C_{1} + λC_{2}) = 0 ……………………………………………………………………………………………………. (‡)

It can be easily seen that (‡) is a line equation. Moreover, it can be proved that **the line with equation g + λh = 0 passes through the point of intersection of g and h**. The proof is as follows.

Let P(x_{0},y_{0}) be the point of intersection of g and h. As a consequence, the following must hold:

A_{1}x_{0} + B_{1}y_{0} + C_{1} = 0 and A_{2}x_{0} + B_{2}y_{0} + C_{2} = 0

(A_{1}x_{0} + B_{1}y_{0} + C_{1}) + λ(A_{2}x_{0} + B_{2}y_{0} + C_{2}) = 0

(A_{1} + λA_{2})x_{0} + (B_{1} + λB_{2})y_{0} + (C_{1} + λC_{2}) = 0 ……………………………………………………………………………………………………………… (‡‡)

From (‡‡), it can be concluded that the line with equation (A_{1} + λA_{2})x + (B_{1} + λB_{2})y + (C_{1} + λC_{2}) = 0 passes through (x_{0},y_{0}). Furthermore, by (‡), the conclusion can be rephrased as: the line with equation g + λh passes through (x_{0},y_{0}). As above is arbitrary, we can say that any line whose equation is of the form g + λh = 0 must pass through the point of intersection of g and h. So, we have proved the following theorem.

**Theorem 1**

Suppose that the lines g and h have the following equations.

g ≡ A_{1}x + B_{1}y + C_{1} = 0

h ≡ A_{2}x + B_{2}y + C_{2} = 0

Let P be the point of intersection of g and h.

For every the line with equation g + λh = 0 passes through P.

We will now prove that **any line that passes through the point of intersection of g and h can be expressed in the form of g + λh = 0** for a . Let the line k ≡ A_{3}x + B_{3}y + C_{3} = 0 pass through the point of intersection of g and h. (k is arbitrary.) We have to show that there exists a such that g + λh = k. Since k ≡ A_{3}x + B_{3}y + C_{3} = 0, the “task” mentioned above is equivalent to the task to prove that there exists a such that g + λh = A_{3}x + B_{3}y + C_{3} = 0. To do this, choose . As g = h = 0, it must hold that g + λh = 0. By the equations of g and h, it follows that:

g + λh = (A_{1} + λA_{2})x + (B_{1} + λB_{2})y + (C_{1} + λC_{2}) = 0 ………………………………………………………………………………………………….. (*)

Substitute the value into (*). As a consequence:

Multiply both sides of the equation by (A_{2}B_{3} – A_{3}B_{2}). Consequently, the equation can be expressed as:

A_{3}(A_{2}B_{1} – A_{1}B_{2})x + B_{3}(A_{2}B_{1} – A_{1}B_{2})y – A_{3}(B_{2}C_{1} – B_{1}C_{2}) – B_{3}(A_{1}C_{2} – A_{2}C_{1}) = 0 ……………………………………………. (**)

Note that the point of intersection of g and h has coordinates (x_{0},y_{0}) with:

……………………………………………………………………………………… (+)

……………………………………………………………………………………. (++)

Since the line k ≡ A_{3}x + B_{3}y + C_{3} = 0 passes through (x_{0},y_{0}), it must hold that:

A_{3}x_{0} + B_{3}y_{0} + C_{3} = 0 ……………………………………………………………………………………………………………………………………………………. (+++)

Substitute (+) and (++) into (+++). We will get the following equation:

– A_{3}(B_{2}C_{1} – B_{1}C_{2}) – B_{3}(A_{1}C_{2} – A_{2}C_{1}) = C_{3}(A_{2}B_{1} – A_{1}B_{2}) ………………………………………………………………………………………… (x)

Note that the last two terms on the left-hand side of (**) are: – A_{3}(B_{2}C_{1} – B_{1}C_{2}) – B_{3}(A_{1}C_{2} – A_{2}C_{1})

They are equal to the terms on the left-hand side of (x). Therefore, we can replace them with the term on the right-hand side of (x). Consequently, we have the following.

A_{3}(A_{2}B_{1} – A_{1}B_{2})x + B_{3}(A_{2}B_{1} – A_{1}B_{2})y + C_{3}(A_{2}B_{1} – A_{1}B_{2}) = 0

Divide both sides of the equation by (A_{2}B_{1} – A_{1}B_{2}). It follows that:

A_{3}x + B_{3}y + C_{3} = 0

We have proved that for every line with equation A_{3}x + B_{3}y + C_{3} = 0 (that passes through the point of intersection of g and h) there is a such that g + λh = A_{3}x + B_{3}y + C_{3} = 0. So, we have proved the following theorem.

**Theorem 2**

Suppose that the lines g and h have the following equations.

g ≡ A_{1}x + B_{1}y + C_{1} = 0

h ≡ A_{2}x + B_{2}y + C_{2} = 0

Let P be the point of intersection of g and h.

Any line passing through P can be expressed in the form of g + λh = 0 for some real number λ.

**Example 1**

Find the equation of the line with slope 1 that passes through the point of intersection of the lines g and h, given:

g ≡ 7x + 5y + 13 = 0

h ≡ 11x – 3y + 37 = 0

**Answer**

By Theorem 2, we can deduce that for some the following holds:

(7 + 11λ)x + (5 – 3λ)y + (13 + 37λ) = 0 ………………………………………………………………………………………………………………………… (o)

Note that (o) is equivalent to:

As the gradient of the line is equal to 1, it follows that

From this we get λ = -1.5. By substituting the value into (o), we obtain the equation of the line in question, i.e. **19x – 19y + 85 = 0**.

**Example 2**

Suppose that the line k passes through (1,-3) and the point of intersection of g and h, where:

g ≡ 71x + 15y + 10 = 0

h ≡ 23x – 17y – 17 = 0

Find the equation of k.

**Answer**

By Theorem 2, the equation of k can be expressed as follows, for some :

(71 + 23λ)x + (15 – 17λ)y + (10 – 17λ) = 0 ……………………………………………………………………………………………………………… (oo)

It is given that k passes through (1,-3). Consequently, x = 1 and y = -3 must satisfy (oo). This results in a linear equation in λ, whose solution is . By substituting this value into (oo), we get the equation of k, that is **1073x + 489y + 394 = 0**.