Straight Lines: Distance and Family of Lines

Distance of a Point from a Line Formula

A fundamental problem in coordinate geometry is determining the shortest distance from a given point to a given straight line. The shortest distance between a point and a line is always measured along the perpendicular segment from the point to the line. The formula for this distance is a powerful tool for solving various geometric problems.

Problem Statement

Given a point $P$ with coordinates $(x_1, y_1)$ and a straight line $L$ defined by the general equation $Ax + By + C = 0$, where $A, B, C$ are real coefficients and $A^2 + B^2 \neq 0$. We want to find the perpendicular distance, denoted by $d$, from the point $P$ to the line $L$.

Let $M$ be the foot of the perpendicular drawn from the point $P$ to the line $L$. The distance we want to find is the length of the segment $PM$, i.e., $d = PM$.

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Formula

The perpendicular distance $d$ from the point $P(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by the formula:

$\mathbf{d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}}$

The numerator $|Ax_1 + By_1 + C|$ is the absolute value of the expression $Ax + By + C$ evaluated at the coordinates of the given point $(x_1, y_1)$. The denominator $\sqrt{A^2 + B^2}$ is the magnitude of the normal vector to the line $(A, B)$. Taking the absolute value in the numerator is essential because distance must be non-negative, while the expression $Ax_1 + By_1 + C$ can be positive, negative, or zero depending on which side of the line the point $(x_1, y_1)$ lies.

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Derivation of the Formula

There are several ways to derive this formula, including using the concept of projections, calculus, or the area of a triangle. Here, we outline a derivation using the area of a triangle, which connects concepts previously discussed.

Let the line $L$ given by $Ax + By + C = 0$ intersect the x-axis and the y-axis at points Q and R respectively. (We assume $A \neq 0$ and $B \neq 0$ for now; special cases will be discussed later).

• To find the x-intercept (point Q), set $y=0$ in the line equation: $Ax + B(0) + C = 0 \implies Ax = -C$. Since $A \neq 0$, $x = -C/A$. So, $Q \equiv (-C/A, 0)$.
• To find the y-intercept (point R), set $x=0$ in the line equation: $A(0) + By + C = 0 \implies By = -C$. Since $B \neq 0$, $y = -C/B$. So, $R \equiv (0, -C/B)$.

Consider the triangle $\triangle PQR$ formed by the given point $P(x_1, y_1)$ and the intercepts $Q(-C/A, 0)$ and $R(0, -C/B)$.

The area of $\triangle PQR$ can be calculated using the area formula from vertex coordinates:

Area($\triangle PQR$) $= \frac{1}{2} | x_1(y_Q - y_R) + x_Q(y_R - y_1) + x_R(y_1 - y_Q) |$

Substitute the coordinates of P, Q, and R:

Simplify equation (i):

Area($\triangle PQR$) $= \frac{1}{2} \left| x_1\left(\frac{C}{B}\right) - \frac{C}{A}\left(-\frac{C}{B} - y_1\right) + 0 \right|$

Area($\triangle PQR$) $= \frac{1}{2} \left| \frac{Cx_1}{B} + \frac{C^2}{AB} + \frac{Cy_1}{A} \right|$

To combine the terms inside the absolute value, find a common denominator, which is $AB$:

Area($\triangle PQR$) $= \frac{1}{2} \left| \frac{Cx_1 \cdot A}{AB} + \frac{C^2}{AB} + \frac{Cy_1 \cdot B}{AB} \right|$

Area($\triangle PQR$) $= \frac{1}{2} \left| \frac{ACx_1 + C^2 + BCy_1}{AB} \right|$

Factor out $C$ from the terms involving $x_1, y_1, C$:

Alternatively, the area of a triangle can be calculated using the formula $\frac{1}{2} \times \text{base} \times \text{height}$. Let the base be the length of the segment $QR$ (the part of line $L$ between the axes). The height corresponding to this base is the perpendicular distance $d$ from point $P$ to the line $L$ (which contains the segment $QR$).

Calculate the length of the base $QR$ using the distance formula for points $Q(-C/A, 0)$ and $R(0, -C/B)$:

$QR = \sqrt{ (0 - (-\frac{C}{A}))^2 + ((-\frac{C}{B}) - 0)^2 }$

$QR = \sqrt{\left(\frac{C}{A}\right)^2 + \left(-\frac{C}{B}\right)^2} = \sqrt{\frac{C^2}{A^2} + \frac{C^2}{B^2}}$

Factor out $C^2$ and find a common denominator inside the square root:

$QR = \sqrt{C^2 \left(\frac{1}{A^2} + \frac{1}{B^2}\right)} = \sqrt{C^2 \left(\frac{B^2 + A^2}{A^2 B^2}\right)}$

Separate the square roots:

$QR = \sqrt{C^2} \times \sqrt{\frac{A^2 + B^2}{A^2 B^2}} = |C| \times \frac{\sqrt{A^2 + B^2}}{\sqrt{A^2 B^2}} = |C| \frac{\sqrt{A^2 + B^2}}{|AB|}$

Now, express the area of $\triangle PQR$ using the base $QR$ and height $d$:

Equate the two expressions for the Area($\triangle PQR$) from equations (ii) and (iv):

Equation (v) can be written as $\frac{1}{2} \frac{|C| |Ax_1 + By_1 + C|}{|AB|} = \frac{1}{2} \frac{|C| \sqrt{A^2 + B^2}}{|AB|} d$.

Assuming $C \neq 0$, we can cancel the common factor $\frac{1}{2} \frac{|C|}{|AB|}$ from both sides:

Solve equation (vi) for $d$:

This derivation assumed $A \neq 0$, $B \neq 0$, and $C \neq 0$. However, the formula obtained is general and holds even for lines parallel to axes ($A=0$ or $B=0$) or lines passing through the origin ($C=0$). The condition $A^2 + B^2 \neq 0$ ensures the denominator is never zero.

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Distance Between Two Parallel Lines

Two distinct straight lines in a plane are parallel if they have the same slope and do not coincide. When two lines are parallel, the perpendicular distance between them is constant throughout their length. We can derive a formula to calculate this distance if the equations of the two parallel lines are known.

Consider two parallel lines $L_1$ and $L_2$. Their general equations can be written such that the coefficients of $x$ and $y$ are identical (or proportional). If they are proportional, we can divide one equation by a constant factor to make the $A$ and $B$ coefficients match. So, let the equations of the two parallel lines be:

where $A, B$ are not both zero ($A^2 + B^2 \neq 0$), and $C_1 \neq C_2$ (if $C_1 = C_2$, the lines would be coincident).

The distance between these two parallel lines is the perpendicular distance from any point on one line to the other line.

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Formula

The perpendicular distance $d$ between the two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ (where $A, B$ are identical in both equations) is given by the formula:

$\mathbf{d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}}$

The absolute value of the difference in the constant terms is used in the numerator, and the denominator is the same normalization factor $\sqrt{A^2 + B^2}$ as in the distance from a point to a line formula.

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Derivation of the Formula

To derive the formula for the distance between the parallel lines $L_1: Ax + By + C_1 = 0$ and $L_2: Ax + By + C_2 = 0$, we can choose any point on one line and calculate its perpendicular distance to the other line. Let's choose an arbitrary point $P(x_1, y_1)$ that lies on the first line, $L_1: Ax + By + C_1 = 0$.

Since the point $P(x_1, y_1)$ lies on the line $L_1$, its coordinates must satisfy the equation of $L_1$:

From equation (3), we can express $Ax_1 + By_1$ as:

The distance $d$ between the parallel lines is the perpendicular distance from the point $P(x_1, y_1)$ (which is on $L_1$) to the second line, $L_2: Ax + By + C_2 = 0$.

Using the formula for the perpendicular distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$, which is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$, we apply this formula with the point $P(x_1, y_1)$ and the line $L_2$ (so $C$ in the formula is $C_2$):

Now, we use the result from equation (4), which states that $Ax_1 + By_1 = -C_1$. Substitute this into the numerator of equation (5):

Simplify the expression inside the absolute value in equation (6):

Since $|C_2 - C_1| = |-(C_1 - C_2)| = |C_1 - C_2|$, the formula can also be written as:

This formula provides the perpendicular distance between the two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$. It assumes that the coefficients of $x$ and $y$ are already identical in both equations. If they are proportional (e.g., $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ where $A_1/A_2 = B_1/B_2 = k \neq 0$), one equation must be multiplied by a suitable constant to make the $A$ and $B$ coefficients match before applying the formula.

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Family of Lines Passing Through the Intersection of Two Given Lines

A family of lines, also called a pencil of lines, is a set of lines that share a common property. One such important family is the set of all lines that pass through the point of intersection of two given distinct lines. This concept is useful for finding the equation of a line that passes through the intersection of two lines without explicitly finding the coordinates of the intersection point.

Let the equations of two intersecting lines be given in their general forms:

Since the lines intersect, there is a unique point $P(x_0, y_0)$ that satisfies both equations simultaneously:

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Equation of the Family of Lines

The equation of any straight line passing through the point of intersection of the two lines $L_1 = 0$ and $L_2 = 0$ can be represented by the linear combination of their equations:

$\mathbf{L_1 + \lambda L_2 = 0}$

or, by substituting the expressions for $L_1$ and $L_2$:

$\mathbf{(A_1 x + B_1 y + C_1) + \lambda (A_2 x + B_2 y + C_2) = 0}$

Here, $\lambda$ (lambda) is a real parameter. By assigning different real values to $\lambda$, we obtain the equations of different lines, all of which pass through the intersection point of $L_1$ and $L_2$.

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Explanation of the Family Equation

Let's verify why this equation represents the family of lines passing through the intersection point:

1. It represents a Straight Line: The equation $(A_1 x + B_1 y + C_1) + \lambda (A_2 x + B_2 y + C_2) = 0$ can be expanded and rearranged into the general form of a linear equation in $x$ and $y$:

   $(A_1 + \lambda A_2)x + (B_1 + \lambda B_2)y + (C_1 + \lambda C_2) = 0$

   This equation is of the form $A'x + B'y + C' = 0$, where $A' = A_1 + \lambda A_2$, $B' = B_1 + \lambda B_2$, and $C' = C_1 + \lambda C_2$. As long as $A'$ and $B'$ are not both zero for a given $\lambda$, this equation represents a straight line.

2. It Passes Through the Intersection Point: Let $P(x_0, y_0)$ be the point of intersection of $L_1 = 0$ and $L_2 = 0$. Substituting the coordinates $(x_0, y_0)$ into the family equation:

   $(A_1 x_0 + B_1 y_0 + C_1) + \lambda (A_2 x_0 + B_2 y_0 + C_2) = 0$

   ... (iii)

   Since $(x_0, y_0)$ is the intersection point, it satisfies both $A_1 x_0 + B_1 y_0 + C_1 = 0$ and $A_2 x_0 + B_2 y_0 + C_2 = 0$. So, substituting these values into equation (iii):

   $(0) + \lambda (0) = 0$

   ($0=0$, always true)

   This confirms that the point $(x_0, y_0)$ satisfies the equation $(A_1 x + B_1 y + C_1) + \lambda (A_2 x + B_2 y + C_2) = 0$ for any real value of $\lambda$. Thus, every line represented by this equation passes through the intersection point of $L_1$ and $L_2$.

3. Parameter $\lambda$ Represents All Lines (with one exception): By changing the value of the parameter $\lambda$, we can obtain the equation of any line that passes through the intersection point of $L_1$ and $L_2$, with one important exception. The equation $L_1 + \lambda L_2 = 0$ can represent any line through the intersection point *except* the line $L_2 = 0$ itself. This is because setting $L_1 + \lambda L_2 = 0$ to be equivalent to $L_2 = 0$ would require the term $L_1$ to vanish, which would imply $L_1 = 0$. This can only happen if the two original lines $L_1$ and $L_2$ were identical, which contradicts the assumption that they are distinct intersecting lines. Informally, to obtain the line $L_2=0$, $\lambda$ would need to be infinitely large relative to the coefficient of $L_1$.

If it is necessary to represent all lines including $L_2=0$, a more general form $\lambda L_1 + \mu L_2 = 0$ can be used, where $\lambda$ and $\mu$ are parameters not both equal to zero. However, $L_1 + \lambda L_2 = 0$ is sufficient in most problems where the task is to find a specific line from the family satisfying an additional condition (unless that condition forces the line to be $L_2$).

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