Orthogonality and equations of lines: 1st grade math course to download in PDF.

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1 I. Directing vectors and Cartesian equation
- 1.1 1. directing vector of a line.
- 1.2 2. Cartesian equation of the right.
2 II. Relative positions of lines

The equation is an important element in mathematics. Its good mastery will allow you to progress throughout the year.

I. Directing vectors and Cartesian equation

In this chapter, we place ourselves in an orthonormal reference frame $(O,\vec{i},\vec{j})$ .

1. directing vector of a line.

Definition:

The directing vector of a line (d) is any representative of the vector $\vec{AB}$ where

A and B are any two distinct points on the line (d).

Example:

In the image below, the vectors $\vec{AB}(2;1)$ , $\vec{u}(-2;-1)$ and $\vec{v}(4;2)$ are vectors

directors of the line (d).

direction vector

Application and Method:

We calculate the coordinates of a directing vector of the line.
The line (BC) and its parallel have the same direction vectors, it is enough to take a representative of origin A.

Example:

Let be three points A(1;5), B(-3;2) and C(2;-1) in an orthonormal frame.

Determine a direction vector of the line (BC).
Detail the construction of the parallel to (BC) passing through A.

2. Cartesian equation of the right.

Theorem:

In an orthonormal reference frame, the coordinates of the set of points M(x;y)

of a line verify a relation where a, b and c are real numbers.

Demonstration:

Let and be two points of a line (d).

Then, for any point belonging to (d), we have :

the vectors $\vec{PM}(x-x_P;y-y_P)$ and $\vec{PQ}(x_Q-x_P;y_Q-y_P)$ are collinear.

So we have $det(\vec{PM};\vec{PQ})=0$ .

That is .

so .

By posing ; and ,

we have the equation of the line (d) which is of the form .

Definition:

The relationship is called a Cartesian equation of the line (d).

Ownership:

The vector $\vec{u}(-b;a)$ is a directing vector of the line (d) of Cartesian equation .

Example:

If the Cartesian equation of the line (d) is , then the vector $\vec{u}(-4;5)$

is a directing vector of this line.

II. Relative positions of lines

1. parallel or secant lines

Theorem:

Let be two lines (d) and (d’) with respective Cartesian equations and where are real numbers.

The lines (d) and (d’) are parallel if, and only if, $ab'-a'b\neq0$ .

Proof:

The director vectors of the lines (d) and (d’) are, respectively, $\vec{u}(-b;a)$ and $\vec{v}(-b';a')$ .

The lines (d) and (d’) are intersecting if and only if the vectors $\vec{u}$ and $\vec{v}$ are not collinear.

In other words, if the determinant of these two vectors is non-zero.

That is, $-ba'-(a(-b'))=-ba'+ab'=ab'-a'b\neq0$ .

2. Secant lines and systems of equations

Theorem:

When two lines intersect, the coordinates (x;y) of their point of intersection

are solution of the system :

system of equations

3. Perpendicular lines

Theorem:

Let be two lines (d) and (d’) with respective Cartesian equations and where are real numbers.

The director vectors of the lines (d) and (d’) are, respectively, $\vec{u}(-b;a)$ and $\vec{v}(-b';a')$ .

The lines (d) and (d’) are perpendicular if, and only if, $\vec{u}.\vec{v}=0$ is .

Proof:

The director vectors of the lines (d) and (d’) are, respectively, $\vec{u}(-b;a)$ and $\vec{v}(-b';a')$ .

The lines are perpendicular if, and only if, these two direction vectors are orthogonal.

This means that the scalar product of these two vectors is zero, i.e. :

$\vec{u}.\vec{v}=0$

This is equivalent to :

$-b\times \,(-b')+a\times \,a'=0$ or .

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