Mathematics

Lines in the Plane

All forms of linear equations, relative positions, angle calculation, parallel/perpendicular lines, and metric problems (distance, perpendicular bisector, angle bisector).

The equation of a line can be determined from a point

P(x_0, y_0)

and a direction vector

\vec{v}(v_x, v_y)

, or by knowing two points

P(x_0, y_0)

and

Q(x_1, y_1)

and finding the vector between them.

The equation of a line can be determined from a point

P(x_0, y_0)

and a direction vector

\vec{v}(v_x, v_y)

, or by knowing two points

P(x_0, y_0)

and

Q(x_1, y_1)

and finding the vector between them.

Finding a line from a point and a vector, or from two points

Guided Example: With

P(3,4)

and

Q(2,2)

we get

\vec{v} = \vec{QP} = (1, 2)

1. Vector Equation:

(x,y) = \lambda(v_x, v_y) + (x_0, y_0)

\color{#2563eb} (x,y) = \lambda(1,2) + (3,4)

2. Parametric Equations: obtained by breaking down the vector equation into its X and Y components.

\begin{cases} x = \lambda \cdot v_x + x_0 \\ y = \lambda \cdot v_y + y_0 \end{cases}

\color{#2563eb} \begin{cases} x = \lambda + 3 \\ y = 2\lambda + 4 \end{cases}

3. Symmetric Equations: (or continuous form). Solve for

\lambda

in both parametric equations and set them equal.

\frac{x - x_0}{v_x} = \frac{y - y_0}{v_y}

\color{#2563eb} \frac{x - 3}{1} = \frac{y - 4}{2}

4. Point-Slope Form: move the denominator

v_x

to the other side to get the slope

m = \frac{v_y}{v_x}

y - y_0 = m(x - x_0)

\color{#2563eb} y - 4 = 2(x - 3)

5. Slope-Intercept Form: solve for

y

in the point-slope equation. Where

n = y_0 - m x_0

is the y-intercept.

y = mx + n

\color{#2563eb} y = 2x - 2

6. General or Standard Form: eliminate denominators and rearrange. With slope

m = -\frac{A}{B}

and direction vector

\vec{v}=(-B, A)

. The normal vector

\vec{n}=(A, B)

is perpendicular to the line.

Ax + By + C = 0

\color{#2563eb} 2x - y - 2 = 0

7. Intercept Form: obtained using the x-intercept,

a

, and the y-intercept,

b

. In our example, the intercepts are

a=1, b=-2

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

\color{#2563eb} \frac{x}{1} + \frac{y}{-2} = 1

Graphical representation of the line and its intercepts a and b with the Cartesian axes

Given a system of two lines in general form

Ax + By + C = 0

and

A'x + B'y + C' = 0

, we can determine their position based on the ratio of their coefficients:

Given a system of two lines in general form

Ax + By + C = 0

and

A'x + B'y + C' = 0

, we can determine their position based on the ratio of their coefficients:

Coincident lines (Infinite solutions):

$\frac{A}{A'} = \frac{B}{B'} = \frac{C}{C'}$

Two lines that exactly overlap (coincident)

Parallel lines (No solution):

$\frac{A}{A'} = \frac{B}{B'} \neq \frac{C}{C'}$

Intersecting lines (1 solution):

$\frac{A}{A'} \neq \frac{B}{B'}$

Two lines intersecting at a single point (secant)

Angle between two lines given their slopes

m_1

and

m_2

\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 \cdot m_2} \right|

Summary of how to construct a line that is parallel or perpendicular to a given line, depending on its equation type:

Original line	Parallel	Perpendicular
$y = mx + n$	$y = mx + n'$	$y = -\frac{1}{m}x + n'$
$Ax + By + C = 0$	$Ax + By + C' = 0$	$Bx - Ay + C' = 0$

Distance from a Point to a Line
The distance from a point

P(x_0, y_0)

to a line

r \equiv Ax + By + C = 0

is:

Distance from a Point to a Line
The distance from a point

P(x_0, y_0)

to a line

r \equiv Ax + By + C = 0

is:

d(P, r) = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}}

Closest Point on a Line
Given a line

r

with direction vector

\vec{v}

and an exterior point

P(x_0, y_0)

, the point

Q(x, y)

on the line that is closest to

P

makes the vector

\vec{PQ}

perpendicular to the line. Its dot product with the direction vector is zero:

\vec{PQ} \cdot \vec{v} = 0

Point Q on line r marking the shortest distance (perpendicular) to exterior point P

Symmetric Point with Respect to a Line (Reflection)
To find the reflection

P'

of point

P

across line

r

First, find the closest point $Q$ on line $r$ from point $P$ .
Then, find the symmetric point $P'$ of $P$ with respect to $Q$ .

Symmetric point P prime across line r, passing through center point Q

Perpendicular Bisector of a Segment
Given a line segment

AB

, the perpendicular bisector is the line that is perpendicular to the segment at its midpoint:

Find the midpoint $M$ of the segment.
Find the line with a direction vector perpendicular to $\vec{AB}$ that passes through $M$ .

Angle Bisector of Two Lines
It is the locus of points

P(x,y)

that are equidistant from two intersecting lines

r

and

s

, such that

d(P, r) = d(P, s)

\frac{|A_1 x + B_1 y + C_1|}{\sqrt{A_1^2 + B_1^2}} = \frac{|A_2 x + B_2 y + C_2|}{\sqrt{A_2^2 + B_2^2}}

(Note: By splitting the absolute value with $\pm$ , you obtain the two angle bisector lines, which are perpendicular to each other).

Two angle bisectors dividing the angles formed by original lines r and s

Lines in the Plane

Equations of a Line

Relative Positions of Two Lines

Constructing Parallel and Perpendicular Lines

Metric Problems