Family of Lines

A family of lines is a collection of lines that satisfy a common property or condition. Each member of the family shares this characteristic, but can vary based on a parameter.

For example, consider the family of lines passing through the origin. All lines in this family have the common property that they intersect the origin \((0, 0)\). The equation of any line passing through the origin can be written as:

\[ y = mx, \]

where \(m\) is the parameter representing the slope of the line. By assigning different values to \(m\), we obtain different lines within this family. Since \(m\) can take any real number value, this family has an infinite number of members. Each value of \(m\) generates a unique line that still satisfies the property of passing through the origin.

In general, a family of lines can be defined by a linear equation that includes a parameter. For instance, the family of lines parallel to a given line can be represented as:

\[ y = mx + c, \]

where \(m\) is fixed (indicating a constant slope) and \(c\) varies. The parameter \(c\) determines the y-intercept of each line. All lines in this family are parallel to each other, sharing the same slope but differing in their vertical displacement.

Another example is the family of lines with a common y-intercept \(b\):

\[ y = mx + b, \]

where \(b\) is fixed and \(m\) varies. Here, all lines intersect the y-axis at the same point \((0, b)\) but have different slopes.

In summary, a family of lines is characterized by a specific property, such as passing through a common point or being parallel. The equation of the family includes a parameter that, when varied, generates the different members of the family. These parameters allow for an infinite number of lines, each satisfying the defined property.

Family of Lines passing through intersection points of two lines

We are particularly interested in the family of lines passing through the intersection point of two non-parallel lines \( L_1 \) and \( L_2 \). Let the equations of these lines be:

\[ L_1: a_1x + b_1y + c_1 = 0 \]

\[ L_2: a_2x + b_2y + c_2 = 0 \]

The family of lines passing through the intersection point of \( L_1 \) and \( L_2 \) is given by the linear combination of these two equations:

\[ L_1 + \lambda L_2 = 0, \]

which can be written as:

\[ (a_1x + b_1y + c_1) + \lambda (a_2x + b_2y + c_2) = 0, \]

or equivalently:

\[ (a_1 + \lambda a_2)x + (b_1 + \lambda b_2)y + (c_1 + \lambda c_2) = 0, \]

where \(\lambda\) is any real number. By varying \(\lambda\), we obtain different members of this family of lines, all of which pass through the intersection point of \(L_1\) and \(L_2\).

For example:

Consider two non-parallel lines:

\[ L_1: 2x + 3y - 6 = 0 \]

\[ L_2: x - y + 1 = 0 \]

We seek the family of lines passing through the intersection point of \(L_1\) and \(L_2\). The family is given by:

\[ L_1 + \lambda L_2 = 0 \]

Substituting the equations of \(L_1\) and \(L_2\):

\[ (2x + 3y - 6) + \lambda (x - y + 1) = 0 \]

Simplifying, we obtain:

\[ (2 + \lambda)x + (3 - \lambda)y + (-6 + \lambda) = 0 \]

This is the general form of the family of lines passing through the intersection of \(L_1\) and \(L_2\). Each value of \(\lambda\) yields a different line in this family.

Proof:

Assume the equations of the lines \( L_1 \) and \( L_2 \) are:

\[ L_1: a_1x + b_1y + c_1 = 0 \]

\[ L_2: a_2x + b_2y + c_2 = 0 \]

Let the point of intersection of \( L_1 \) and \( L_2 \) be \((\alpha, \beta)\). Since \((\alpha, \beta)\) lies on both lines, it satisfies both equations. Thus,

\[ a_1\alpha + b_1\beta + c_1 = 0 \quad \text{(1)} \]

\[ a_2\alpha + b_2\beta + c_2 = 0 \quad \text{(2)} \]

Consider the linear combination of \( L_1 \) and \( L_2 \):

\[ (a_1x + b_1y + c_1) + \lambda (a_2x + b_2y + c_2) = 0 \]

Simplifying, we get:

\[ (a_1 + \lambda a_2)x + (b_1 + \lambda b_2)y + (c_1 + \lambda c_2) = 0 \]

Since this equation is of the form \( ax + by + c = 0 \), it represents a straight line.

Next, substitute the point \((\alpha, \beta)\) into this equation:

\[ (a_1\alpha + b_1\beta + c_1) + \lambda (a_2\alpha + b_2\beta + c_2) = 0 \]

Substitute the values from (1) and (2):

\[ 0 + \lambda \cdot 0 = 0 \]

This confirms that \((\alpha, \beta)\) satisfies the equation, proving that it is a line passing through the intersection point of \( L_1 \) and \( L_2 \).

As \(\lambda\) varies over all real numbers, we obtain different lines passing through the intersection point, forming the family of lines. Thus, the family of lines passing through the intersection point of \( L_1 \) and \( L_2 \) is given by:

\[ L_1 + \lambda L_2 = 0. \]

Example

Find the equation of the line passing through the intersection of the lines \(3x + 11y = 1\) and \(x + 4y = 2\) and also passing through the origin.

Solution:

Since the line passes through the intersection of the given lines, its equation can be assumed to be:

\[ 3x + 11y - 1 + \lambda (x + 4y - 2) = 0 \]

for some \(\lambda \in \mathbb{R}\).

We need to determine the value of \(\lambda\). Given that the line passes through the origin, the point \((0, 0)\) must satisfy this equation. Substituting \((0, 0)\) into the equation:

\[ 3(0) + 11(0) - 1 + \lambda (0 + 4(0) - 2) = 0 \]

Simplifies to:

\[ -1 + \lambda (-2) = 0 \]

Solving for \(\lambda\):

\[ -1 - 2\lambda = 0 \implies \lambda = -\frac{1}{2} \]

Now, substitute \(\lambda = -\frac{1}{2}\) back into the assumed equation:

\[ 3x + 11y - 1 + \left(-\frac{1}{2}\right) (x + 4y - 2) = 0 \]

Simplify the equation:

\[ 3x + 11y - 1 - \frac{1}{2}(x + 4y - 2) = 0 \]

Distribute the \(-\frac{1}{2}\):

\[ 3x + 11y - 1 - \frac{1}{2}x - 2y + 1 = 0 \]

Combine like terms:

\[ \left(3 - \frac{1}{2}\right)x + (11 - 2)y = 0 \]

\[ \frac{5}{2}x + 9y = 0 \implies 5x + 18y = 0 \]

Thus, the equation of the line passing through the intersection of \(3x + 11y = 1\) and \(x + 4y = 2\) and also passing through the origin is:

\[ 5x + 18y = 0. \]

Why is this family of lines so important?

In the above example, we found the equation of the line passing through the intersection of the lines \(3x + 11y = 1\) and \(x + 4y = 2\) without directly calculating the intersection point. This method is significant because it leverages the family of lines passing through the intersection point, enabling us to derive the desired line equation efficiently. Thus, the family of lines is a powerful tool for solving such problems by bypassing the need for explicit intersection point computation.

Example

Consider three lines \( L_1: 2x + y = 5 \), \( L_2: y = x - 1 \), and \( L_3: y = 4x + 2 \). \( L_2 \) and \( L_3 \) intersect at \( A \), \( L_3 \) and \( L_1 \) intersect at \( B \), and \( L_1 \) and \( L_2 \) intersect at \( C \). Find the equation of the altitude dropped from vertex \( A \).

Solution:

Since the altitude passes through the intersection of \( L_2 \) and \( L_3 \) (point \( A \)), its equation can be assumed to be:

\[ L_2 + \lambda L_3 = 0 \]

That is,

\[ (x - y - 1) + \lambda (4x - y + 2) = 0 \]

Simplifying, we get:

\[ (1 + 4\lambda)x - (1 + \lambda)y + (2\lambda - 1) = 0 \]

The slope of this line is:

\[ \frac{1 + 4\lambda}{1 + \lambda} \]

The slope of \( BC \), that is, \( L_3 \), is 4. Since the altitude is perpendicular to \( BC \):

\[ \frac{1 + 4\lambda}{1 + \lambda} \cdot 4 = -1 \]

Solving for \(\lambda\):

\[ 4 + 16\lambda = -1 - \lambda \]

\[ 17\lambda = -5 \]

\[ \lambda = -\frac{5}{17} \]

Substituting \(\lambda = -\frac{5}{17}\) into the equation:

\[ x - y - 1 - \frac{5}{17}(4x - y + 2) = 0 \]

Simplifying,

\[ 17x - 17y - 17 - 20x + 5y - 10 = 0 \]

\[ -3x - 12y - 27 = 0 \]

\[ x + 4y + 9 = 0 \]

Thus, the equation of the altitude dropped from vertex \( A \) is:

\[ x + 4y + 9 = 0 \]