Straight Lines: General Equation and Related Concepts

General Equation of a Line ($Ax+By+C=0$)

In coordinate geometry, straight lines are represented by linear equations involving the variables $x$ and $y$. The most comprehensive and universally applicable form for the equation of a straight line in two dimensions is the General Form.

Any equation of the first degree (meaning the highest power of $x$ and $y$ is 1) in two variables $x$ and $y$ can be written in the form:

$\mathbf{Ax + By + C = 0}$

Here, $A$, $B$, and $C$ are real constant coefficients. The fundamental condition for this equation to represent a straight line is that the coefficients $A$ and $B$ are not both simultaneously zero. This is often expressed as $A^2 + B^2 \neq 0$.

Characteristics of the General Equation

• Universality: The greatest strength of the general form $Ax + By + C = 0$ is that it can represent every possible straight line in the Cartesian plane. Unlike some other forms (like slope-intercept form or intercept form) which may not represent vertical lines or lines through the origin in all cases, the general form is always applicable.

• Role of Constants: The specific values of the constants $A$, $B$, and $C$ determine the unique position and orientation of the line in the plane.

  • The coefficients $A$ and $B$ are related to the slope or direction of the line. As we've seen, if $B \neq 0$, the slope $m = -\frac{A}{B}$. If $B = 0$, the line is vertical.
  • The constant term $C$ influences the position of the line relative to the origin and axes, affecting its intercepts.

• Non-Uniqueness: It is important to note that the general form of the equation for a given line is not unique. If $Ax + By + C = 0$ represents a line, then multiplying the entire equation by any non-zero real constant $k$ will result in an equivalent equation that represents the exact same line:

  $k(Ax + By + C) = k \cdot 0$

  ... (i)

  $(kA)x + (kB)y + (kC) = 0$

  ... (ii)

  This means, for instance, $2x + 3y - 5 = 0$ and $4x + 6y - 10 = 0$ represent the same line.

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Special Cases based on Coefficients

The values of $A$ and $B$ (and sometimes $C$) determine the specific orientation of the line represented by $Ax + By + C = 0$:

• If $A = 0$ (and $B \neq 0$):

  The equation reduces to $0 \cdot x + By + C = 0$, which is $By + C = 0$. Since $B \neq 0$, we can solve for $y$: $y = -\frac{C}{B}$. This is the equation of a horizontal line that is parallel to the x-axis. Its slope is 0.

• If $B = 0$ (and $A \neq 0$):

  The equation reduces to $Ax + 0 \cdot y + C = 0$, which is $Ax + C = 0$. Since $A \neq 0$, we can solve for $x$: $x = -\frac{C}{A}$. This is the equation of a vertical line that is parallel to the y-axis. Its slope is undefined.

• If $C = 0$:

  The equation becomes $Ax + By + 0 = 0$, or $Ax + By = 0$. For any point $(x, y) = (0, 0)$, we have $A(0) + B(0) = 0$, which is true. This means the point $(0, 0)$ (the origin) always satisfies the equation $Ax + By = 0$. Therefore, any line with its constant term equal to zero passes through the origin.

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The Condition $A^2 + B^2 \neq 0$

The requirement that $A$ and $B$ are not simultaneously zero ($A^2 + B^2 \neq 0$) is crucial for the equation $Ax + By + C = 0$ to represent a straight line.

Consider what happens if both $A=0$ and $B=0$:

• If $A = 0$, $B = 0$, and $C = 0$:

  The equation becomes $0 \cdot x + 0 \cdot y + 0 = 0$, which simplifies to $0 = 0$. This is an identity that is true for any pair of values $(x, y)$. Thus, the equation $0x + 0y + 0 = 0$ represents the entire Cartesian plane, not just a single line.

• If $A = 0$, $B = 0$, and $C \neq 0$:

  The equation becomes $0 \cdot x + 0 \cdot y + C = 0$, which simplifies to $C = 0$. Since we assumed $C$ is a non-zero constant, this is a contradiction (e.g., $5 = 0$). This means there are no points $(x, y)$ that satisfy the equation. This represents an empty set, not a line.

In both cases where $A=0$ and $B=0$, the equation does not represent a straight line. Hence, the condition $A^2 + B^2 \neq 0$ (which means at least one of $A$ or $B$ is non-zero) is a necessary part of the definition of the general form of a linear equation representing a straight line.

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Converting General Equation to Other Standard Forms

The General Equation of a Line, $Ax + By + C = 0$ (where $A^2 + B^2 \neq 0$), is universal in its ability to represent any straight line in the Cartesian plane. While it is comprehensive, other standard forms of linear equations are often more convenient for directly extracting specific geometric information about the line, such as its slope, intercepts, or distance from the origin.

It is therefore important to know how to convert the general equation into these other forms.

1. Conversion to Slope-Intercept Form ($y = mx + c$)

The Slope-Intercept Form directly provides the slope ($m$) and the y-intercept ($c$) of a line. This conversion is possible for any line that is not vertical.

Condition for Applicability:

This conversion is applicable only if the line is not vertical, which means the coefficient of $y$ in the general equation cannot be zero. Thus, $B \neq 0$.

Conversion Steps:

Start with the general equation: $Ax + By + C = 0$.

Our goal is to isolate the $y$ term on one side of the equation.

Since $B \neq 0$, we can divide the entire equation by $B$:

Simplify equation (i):

Equation (ii) is in the Slope-Intercept Form $y = mx + c$.

Parameters from General Form:

By comparing equation (ii) with the standard form $y = mx + c$, we can identify the slope $m$ and the y-intercept $c$ in terms of the coefficients $A$, $B$, and $C$ from the general equation:

• Slope: $\mathbf{m = -\frac{A}{B}}$
• Y-intercept: $\mathbf{c = -\frac{C}{B}}$

These formulas are valid for any non-vertical line $Ax + By + C = 0$ (where $B \neq 0$).

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Answer:

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2. Conversion to Intercept Form ($\frac{x}{a} + \frac{y}{b} = 1$)

The Intercept Form directly provides the x-intercept ($a$) and the y-intercept ($b$) of a line. This conversion is possible for lines that do not pass through the origin and are not parallel to either axis.

Conditions for Applicability:

This conversion is applicable only if the line is not parallel to the x-axis ($A \neq 0$), not parallel to the y-axis ($B \neq 0$), and does not pass through the origin ($C \neq 0$). So, $A \neq 0$, $B \neq 0$, and $C \neq 0$.

Conversion Steps:

Start with the general equation: $Ax + By + C = 0$.

Move the constant term to the right-hand side:

Since $C \neq 0$, the right-hand side $-C$ is non-zero. Divide the entire equation by $-C$ to make the right-hand side equal to 1:

Simplify equation (iv) and rewrite it in the standard intercept form $\frac{x}{a} + \frac{y}{b} = 1$ by putting the denominators explicitly under $x$ and $y$ terms:

Equation (v) is the Intercept Form.

Parameters from General Form:

By comparing equation (v) with the standard form $\frac{x}{a} + \frac{y}{b} = 1$, we can identify the x-intercept $a$ and the y-intercept $b$ in terms of the coefficients $A$, $B$, and $C$ from the general equation:

• X-intercept: $\mathbf{a = -\frac{C}{A}}$
• Y-intercept: $\mathbf{b = -\frac{C}{B}}$

These formulas are valid for lines where $A \neq 0$, $B \neq 0$, and $C \neq 0$. Note that the y-intercept formula $b = -C/B$ is the same as $c = -C/B$ from the slope-intercept form, as they represent the same quantity.

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Answer:

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3. Conversion to Normal Form ($x \cos \omega + y \sin \omega = p$)

The Normal Form provides the perpendicular distance ($p$) from the origin to the line and the angle ($\omega$) that this perpendicular makes with the positive x-axis. This conversion is possible for any line (except possibly one passing through the origin, where $p=0$).

Condition for Applicability:

This conversion is applicable for any line, provided $A$ and $B$ are not both zero ($A^2 + B^2 \neq 0$). If $C=0$, then $p=0$, and $\omega$ is related to the slope/direction of the line itself.

Conversion Steps:

Start with the general equation: $Ax + By + C = 0$.

Move the constant term to the right-hand side: $Ax + By = -C$.

In the normal form $x \cos \omega + y \sin \omega = p$, the distance $p$ is non-negative. We need to ensure the right-hand side of our equation is non-negative. Let the equation be $A'x + B'y = D$, where $D = |-C| = \sqrt{C^2}$.

• If $-C \ge 0$ (i.e., $C \le 0$), we use $A'=A$, $B'=B$, $D=-C$.
• If $-C < 0$ (i.e., $C > 0$), we multiply the equation $Ax + By = -C$ by -1, getting $-Ax - By = C$. We use $A'=-A$, $B'=-B$, $D=C$.

In both cases, $\sqrt{(A')^2 + (B')^2} = \sqrt{(\pm A)^2 + (\pm B)^2} = \sqrt{A^2 + B^2}$.

The coefficients of $x$ and $y$ in the normal form ($ \cos \omega$ and $ \sin \omega$) have the property that the sum of their squares is 1 ($\cos^2 \omega + \sin^2 \omega = 1$). To make the coefficients in $A'x + B'y = D$ satisfy this property, we divide the entire equation by $\sqrt{A'^2 + B'^2}$, which is equal to $\sqrt{A^2 + B^2}$. This is the normalization step.

Equation (vi) is in the Normal Form $x \cos \omega + y \sin \omega = p$.

Parameters from General Form:

By comparing equation (vi) with the standard form $x \cos \omega + y \sin \omega = p$, we get:

• Perpendicular distance from origin: $\mathbf{p = \frac{D}{\sqrt{A^2 + B^2}} = \frac{|-C|}{\sqrt{A^2 + B^2}} = \frac{|C|}{\sqrt{A^2 + B^2}}}$
• $\mathbf{\cos \omega = \frac{A'}{\sqrt{A^2 + B^2}}}$
• $\mathbf{\sin \omega = \frac{B'}{\sqrt{A^2 + B^2}}}$

Remember that $A'$ and $B'$ are $A$ and $B$ (if $C \le 0$) or $-A$ and $-B$ (if $C > 0$), chosen so that the constant term on the RHS is non-negative before dividing by the square root.

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Answer:

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Summary of Conversions from $Ax + By + C = 0$

Here's a summary of how to convert the general equation $Ax + By + C = 0$ (where $A^2 + B^2 \neq 0$) to other standard forms:

Target Form	Conversion Procedure	Parameters Found	Conditions
Slope-Intercept ($y = mx + c$)	Isolate $y$: $By = -Ax - C$, then divide by $B$: $\mathbf{y = (-\frac{A}{B})x + (-\frac{C}{B})}$	$m = -A/B$ $c = -C/B$	$B \neq 0$
Intercept ($\frac{x}{a} + \frac{y}{b} = 1$)	Move constant to RHS: $Ax + By = -C$, then divide by $-C$: $\frac{Ax}{-C} + \frac{By}{-C} = 1 \implies \mathbf{\frac{x}{(-C/A)} + \frac{y}{(-C/B)} = 1}$	$a = -C/A$ $b = -C/B$	$A \neq 0, B \neq 0, C \neq 0$
Normal ($x \cos \omega + y \sin \omega = p$)	Rewrite as $Ax + By = -C$. If $-C < 0$, multiply by -1 to get $A'x + B'y = D$ (where $D \ge 0$). Divide by $\sqrt{A^2+B^2}$: $\mathbf{\frac{A'}{\sqrt{A^2+B^2}} x + \frac{B'}{\sqrt{A^2+B^2}} y = \frac{D}{\sqrt{A^2+B^2}}}$	$p = \frac{D}{\sqrt{A^2+B^2}} = \frac{|C|}{\sqrt{A^2+B^2}}$ $\cos \omega = \frac{A'}{\sqrt{A^2+B^2}}$ $\sin \omega = \frac{B'}{\sqrt{A^2+B^2}}$	$A^2+B^2 \neq 0$

Note: $A'$ and $B'$ in the Normal Form conversion refer to the coefficients after adjusting the sign of the constant term to make the RHS non-negative.

Point of Intersection of Two Lines

When two distinct straight lines are drawn in a plane, there are two possibilities for their relative positions: they are either parallel and never intersect, or they intersect at exactly one point. If the lines are identical (coincident), they share infinitely many points.

The point of intersection of two distinct, non-parallel lines is the single point that lies on both lines simultaneously. Because this point is common to both lines, its coordinates must satisfy the algebraic equations that represent each line.

Let the equations of the two lines be given in the general form:

where $A_1, B_1, C_1, A_2, B_2, C_2$ are real constants, and $A_1^2 + B_1^2 \neq 0$, $A_2^2 + B_2^2 \neq 0$.

To find the coordinates of the point of intersection $(x_0, y_0)$, we need to find the values of $x$ and $y$ that satisfy both equation (1) and equation (2) simultaneously. This is a classic problem of solving a system of two linear equations in two variables.

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Methods for Solving the System

There are several standard algebraic methods to solve a system of two linear equations:

1. Substitution Method:

   • Solve one of the equations for one variable in terms of the other variable (e.g., solve equation (1) for $y$ in terms of $x$, assuming $B_1 \neq 0$).
   • Substitute the expression obtained in the previous step into the other equation (equation (2)).
   • This will result in a single linear equation in one variable ($x$). Solve this equation to find the value of that variable.
   • Substitute the value found back into the expression obtained in the first step to find the value of the other variable ($y$).

2. Elimination Method:

   • Multiply one or both equations by suitable non-zero constants such that the coefficients of one of the variables (either $x$ or $y$) become equal in magnitude. You might choose to make them have opposite signs for easier addition.
   • Add or subtract the modified equations to eliminate one variable.
   • Solve the resulting single-variable equation.
   • Substitute the value obtained back into either of the original equations to find the value of the eliminated variable.

3. Cross-Multiplication Method (or Cramer's Rule for 2x2 systems):

   This method provides direct formulas for the values of $x$ and $y$ in terms of the coefficients, provided a unique solution exists. For the system:

   $A_1 x + B_1 y + C_1 = 0$

   $A_2 x + B_2 y + C_2 = 0$

   Arrange the coefficients and constants as follows and apply the cross-multiplication rule:

   $\frac{x}{B_1 C_2 - B_2 C_1} = \frac{y}{C_1 A_2 - C_2 A_1} = \frac{1}{A_1 B_2 - A_2 B_1}$

   From this relationship, we can write the explicit formulas for $x$ and $y$:

   $x = \frac{B_1 C_2 - B_2 C_1}{A_1 B_2 - A_2 B_1}$

   ... (3)

   $y = \frac{C_1 A_2 - C_2 A_1}{A_1 B_2 - A_2 B_1}$

   ... (4)

   This method yields a unique solution if and only if the denominator $A_1 B_2 - A_2 B_1 \neq 0$. The term $A_1 B_2 - A_2 B_1$ is the determinant of the coefficient matrix $\begin{pmatrix} A_1 & B_1 \\ A_2 & B_2 \end{pmatrix}$.

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Conditions for Intersection, Parallelism, and Coincidence

The geometric relationship between two lines can be determined by examining the ratios of their corresponding coefficients in the general form $A_1 x + B_1 y + C_1 = 0$ and $A_2 x + B_2 y + C_2 = 0$.

1. Intersecting Lines (Unique Solution):

   The two lines intersect at exactly one point if and only if their slopes are different. The slope of $L_1$ is $m_1 = -A_1/B_1$ (if $B_1 \neq 0$) and the slope of $L_2$ is $m_2 = -A_2/B_2$ (if $B_2 \neq 0$).

   Condition $m_1 \neq m_2 \implies -\frac{A_1}{B_1} \neq -\frac{A_2}{B_2} \implies \frac{A_1}{B_1} \neq \frac{A_2}{B_2}$.

   Cross-multiplying (if $B_1, B_2 \neq 0$) gives $A_1 B_2 \neq A_2 B_1$, which can be written as $A_1 B_2 - A_2 B_1 \neq 0$. This is the same condition for the denominator in the cross-multiplication method to be non-zero.

   This condition is most generally expressed as:

   $\mathbf{\frac{A_1}{A_2} \neq \frac{B_1}{B_2}}$

   provided $A_2, B_2$ are non-zero. If $A_2=0$ or $B_2=0$, the condition needs careful handling (e.g., if $A_2=0$, the condition becomes $A_1 B_2 - 0 \cdot B_1 \neq 0 \implies A_1 B_2 \neq 0$, meaning $A_1 \neq 0$ and $B_2 \neq 0$, which corresponds to a vertical line intersecting a non-horizontal line). The determinant condition $A_1 B_2 - A_2 B_1 \neq 0$ is universally applicable.

2. Parallel and Distinct Lines (No Solution):

   The two lines are parallel but do not overlap if they have the same slope but different y-intercepts (assuming non-vertical). Condition $m_1 = m_2 \implies \frac{A_1}{A_2} = \frac{B_1}{B_2}$. Condition for different y-intercepts $c_1 \neq c_2 \implies -\frac{C_1}{B_1} \neq -\frac{C_2}{B_2} \implies \frac{C_1}{B_1} \neq \frac{C_2}{B_2}$ (if $B_1, B_2 \neq 0$). The complete condition for parallel and distinct lines is that the ratio of x-coefficients equals the ratio of y-coefficients, but this is not equal to the ratio of the constant terms:

   $\mathbf{\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}}$

   This applies when $A_2, B_2, C_2$ are non-zero. If any of the denominators are zero, the ratios must be interpreted carefully (e.g., $A_1/0 = A_2/0$ implies $A_1=0, A_2=0$, corresponding to horizontal lines). The condition $A_1 B_2 - A_2 B_1 = 0$ and $B_1 C_2 - B_2 C_1 \neq 0$ (or $C_1 A_2 - C_2 A_1 \neq 0$) is universally applicable.

3. Coincident Lines (Infinitely Many Solutions):

   The two lines are identical if they have the same slope and the same y-intercept. This happens when the equation of one line is a non-zero constant multiple of the other equation. The ratio of all corresponding coefficients is equal:

   $\mathbf{\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}}$

   This applies when $A_2, B_2, C_2$ are non-zero. This means $A_1 B_2 - A_2 B_1 = 0$, $B_1 C_2 - B_2 C_1 = 0$, and $C_1 A_2 - C_2 A_1 = 0$. If the equations are proportional, any solution to one equation is a solution to the other, leading to infinitely many solutions (all the points on the line).

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Answer:

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