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 A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0, respectively.

g ≡ A1x + B1y + C1 = 0 or g = 0

h ≡ A2x + B2y + C2 = 0 or h = 0

Let \lambda \in \mathbb{R}. It can be easily proved that g + λh = 0 is a line equation.

By substitution, we get the following:

g + λh = (A1 + λA2)x + (B1 + λB2)y + (C1 + λC2) = 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(x0,y0) be the point of intersection of g and h. As a consequence, the following must hold:

A1x0 + B1y0 + C1 = 0 and A2x0 + B2y0 + C2 = 0

(A1x0 + B1y0 + C1) + λ(A2x0 + B2y0 + C2) = 0

(A1 + λA2)x0 + (B1 + λB2)y0 + (C1 + λC2) = 0     ……………………………………………………………………………………………………………… (‡‡)

From (‡‡), it can be concluded that the line with equation (A1 + λA2)x + (B1 + λB2)y + (C1 + λC2) = 0 passes through (x0,y0). Furthermore, by (‡), the conclusion can be rephrased as: the line with equation g + λh passes through (x0,y0). As \lambda \in \mathbb{R} 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 ≡ A1x + B1y + C1 = 0

h ≡ A2x + B2y + C2 = 0

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

For every \lambda \in \mathbb{R} 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 \lambda \in \mathbb{R}. Let the line k ≡ A3x + B3y + C3 = 0 pass through the point of intersection of g and h. (k is arbitrary.) We have to show that there exists a \lambda \in \mathbb{R} such that g + λh = k. Since k ≡ A3x + B3y + C3 = 0, the “task” mentioned above is equivalent to the task to prove that there exists a \lambda \in \mathbb{R} such that g + λh = A3x + B3y + C3 = 0. To do this, choose \lambda = \frac{A_3 B_1 - A_1 B_3}{A_2 B_3 - A_3 B_2}. As g = h = 0, it must hold that g + λh = 0. By the equations of g and h, it follows that:

g + λh = (A1 + λA2)x + (B1 + λB2)y + (C1 + λC2) = 0 ………………………………………………………………………………………………….. (*)

Substitute the value \lambda = \frac{A_3 B_1 - A_1 B_3}{A_2 B_3 - A_3 B_2} into (*). As a consequence:

(A_1 + \frac{A_3 B_1 - A_1 B_3}{A_2 B_3 - A_3 B_2} A_2) x + (B_1 + \frac{A_3 B_1 - A_1 B_3}{A_2 B_3 - A_3 B_2} B_2) y + ( C_1 + \frac{A_3 B_1 - A_1 B_3}{A_2 B_3 - A_3 B_2} C_2) = 0

Multiply both sides of the equation by (A2B3 – A3B2). Consequently, the equation can be expressed as:

A3(A2B1 – A1B2)x + B3(A2B1 – A1B2)y – A3(B2C1 – B1C2) – B3(A1C2 – A2C1) = 0 ……………………………………………. (**)

Note that the point of intersection of g and h has coordinates (x0,y0) with:

x_0 = \frac{B_1 C_2 - B_2 C_1}{A_1 B_2 - A_2 B_1} ……………………………………………………………………………………… (+)

y_0 = \frac{A_2 C_1 - A_1 C_2}{A_1 B_2 - A_2 B_1} ……………………………………………………………………………………. (++)

Since the line k ≡ A3x + B3y + C3 = 0 passes through (x0,y0), it must hold that:

A3x0 + B3y0 + C3 = 0 ……………………………………………………………………………………………………………………………………………………. (+++)

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

– A3(B2C1 – B1C2) – B3(A1C2 – A2C1) = C3(A2B1 – A1B2) ………………………………………………………………………………………… (x)

Note that the last two terms on the left-hand side of (**) are:  – A3(B2C1 – B1C2) – B3(A1C2 – A2C1)

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.

A3(A2B1 – A1B2)x + B3(A2B1 – A1B2)y + C3(A2B1 – A1B2) = 0

Divide both sides of the equation by (A2B1 – A1B2). It follows that:

A3x + B3y + C3 = 0

We have proved that for every line with equation A3x + B3y + C3 = 0 (that passes through the point of intersection of g and h) there is a \lambda \in \mathbb{R} such that g + λh = A3x + B3y + C3 = 0. So, we have proved the following theorem.

 

Theorem 2

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

g ≡ A1x + B1y + C1 = 0

h ≡ A2x + B2y + C2 = 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 \lambda \in \mathbb{R} the following holds:

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

Note that (o) is equivalent to:

y = \frac{7 + 11 \lambda}{3 \lambda - 5} x + \frac{13 + 37 \lambda}{3 \lambda - 5}

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

\frac{7 + 11 \lambda}{3 \lambda - 5} = 1

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 \lambda \in \mathbb{R}:

(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 \lambda = - \frac{12}{19}. By substituting this value into (oo), we get the equation of k, that is 1073x + 489y + 394 = 0.

 

 

 

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