The Straight Line

A straight line is the locus of all points in the plane that follow a constant direction.

The angle of inclination $\theta$ of a line is the angle measured counterclockwise from the positive $x$-axis to the line, such that $0^\circ \leq \theta < 180^\circ$.

The slope $m$ of a line is defined as the tangent of its angle of inclination:

$$\boxed{m = \tan \theta}$$

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Given a point $P_1(x_1, y_1)$ and a slope $m$:

$$y - y_1 = m(x - x_1)$$

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Given two distinct points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, with $x_1 \ne x_2$:

$$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$

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Given slope $m$ and $y$-intercept $b$ (the point where the line crosses the $y$-axis):

$$y = mx + b$$

Where:

• $m$: slope
• $b$: $y$-intercept

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If the line intersects the $x$-axis at $(a, 0)$ and the $y$-axis at $(0, b)$, with $a \ne 0$ and $b \ne 0$:

$$\frac{x}{a} + \frac{y}{b} = 1$$

Where:

• $a$: $x$-intercept
• $b$: $y$-intercept

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Any straight line can be written as:

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

where $A, B, C \in \mathbb{R}$, and not all are zero.

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Given two lines with slopes $m_1$ and $m_2$, the acute angle $\theta$ between them is:

$$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$$

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Consider the lines:

$$\begin{aligned} \mathscr{L}_1 &: A_1x + B_1y + C_1 = 0 \\ \mathscr{L}_2 &: A_2x + B_2y + C_2 = 0 \end{aligned}$$

Two lines are perpendicular if their slopes satisfy $m_1 \cdot m_2 = -1$. In terms of coefficients:

$$\boxed{A_1 A_2 + B_1 B_2 = 0}$$

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Two lines are parallel if they have equal slopes:

$$m_1 = m_2 \quad \Rightarrow \quad \boxed{\frac{A_1}{A_2} = \frac{B_1}{B_2}} \quad \text{(provided } A_2, B_2 \ne 0\text{)}$$

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Two lines are coincident if all their coefficients are proportional:

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

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Two lines intersect at exactly one point if they are not parallel:

$$\boxed{A_1 B_2 - A_2 B_1 \ne 0}$$

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The normal form of a line is:

$$x \cos \omega + y \sin \omega - p = 0$$

Where:

• $\omega$: angle between the normal vector and the positive $x$-axis ($0^\circ \leq \omega < 360^\circ$)
• $p$: perpendicular distance from the origin to the line (always $p \geq 0$)

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Given $Ax + By + C = 0$, divide by $\sqrt{A^2 + B^2}$, choosing the sign opposite to that of $C$ to ensure $p \geq 0$:

$$\frac{A}{\pm \sqrt{A^2 + B^2}}x + \frac{B}{\pm \sqrt{A^2 + B^2}}y + \frac{C}{\pm \sqrt{A^2 + B^2}} = 0$$

The sign is chosen so that $\dfrac{C}{\pm \sqrt{A^2 + B^2}} \leq 0$, which guarantees $p = -\dfrac{C}{\pm \sqrt{A^2 + B^2}} \geq 0$.

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Given a point $P_1(x_1, y_1)$ and a line $Ax + By + C = 0$, the (always non-negative) perpendicular distance is:

$$d(P_1, \mathscr{L}) = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

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The directed distance $d$ carries a sign that depends on the orientation of the normal vector $(A, B)$:

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

1. Line not passing through the origin ($C \ne 0$):

   • $d > 0$ if $P_1$ and the origin lie on opposite sides of the line.
   • $d < 0$ if they lie on the same side.

   

2. Line passing through the origin ($C = 0$):

   • $d > 0$ if $P_1$ is "above" the line (in the direction of $(A, B)$).
   • $d < 0$ if it is "below".

   

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Given two lines $\mathscr{L}_1: A_1x + B_1y + C_1 = 0$ and $\mathscr{L}_2: A_2x + B_2y + C_2 = 0$, the angle bisectors are the loci of points equidistant to both lines:

$$\frac{|A_1x + B_1y + C_1|}{\sqrt{A_1^2 + B_1^2}} = \frac{|A_2x + B_2y + C_2|}{\sqrt{A_2^2 + B_2^2}}$$

Removing absolute values gives the two bisectors:

$$\frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}}$$

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Given two parallel lines $\mathscr{L}_1: Ax + By + C_1 = 0$ and $\mathscr{L}_2: Ax + By + C_2 = 0$ (same $A$ and $B$), the distance between them is:

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

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Given three vertices $P_1(x_1, y_1)$, $P_2(x_2, y_2)$, $P_3(x_3, y_3)$, the area of the triangle is:

$$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

Or using a determinant:

$$\text{Area} = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} \right|$$

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Given $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, the equation of the line is:

$$\begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ \end{vmatrix} = 0$$

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Given $Ax + By + C = 0$, the family of parallel lines is:

$$Ax + By + k = 0 \quad \text{for } k \in \mathbb{R}$$

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If a given line has slope $m$, all perpendicular lines have slope $-\dfrac{1}{m}$. If they pass through a fixed point $(x_0, y_0)$:

$$y - y_0 = -\frac{1}{m}(x - x_0)$$

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Given two intersecting lines $\mathscr{L}_1: A_1x + B_1y + C_1 = 0$ and $\mathscr{L}_2: A_2x + B_2y + C_2 = 0$, the family of all lines passing through their intersection point is:

$$A_1x + B_1y + C_1 + \lambda(A_2x + B_2y + C_2) = 0 \quad \text{for } \lambda \in \mathbb{R}$$