Vector coordinates: AB = (xB − xA, yB − yA)

The formula to know

In a coordinate frame, if $A(x_A ; y_A)$ and $B(x_B ; y_B)$:

$\overrightarrow{AB} = \begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}$

Memory hint: “arrival minus departure”.

Example

$A(2 ; 5)$ and $B(7 ; 3)$.

$\overrightarrow{AB} = (7 - 2 ; 3 - 5) = (5 ; -2)$.

Operations on coordinates

If $\vec{u} = (a ; b)$ and $\vec{v} = (c ; d)$:

• Sum: $\vec{u} + \vec{v} = (a + c ; b + d)$
• Scalar multiple: $k \vec{u} = (k a ; k b)$
• Opposite: $-\vec{u} = (-a ; -b)$

Midpoint of a segment

$M$ midpoint of $[AB]$:

$M = \left( \dfrac{x_A + x_B}{2} \, ; \, \dfrac{y_A + y_B}{2} \right)$

Memory hint: “average of coordinates”.

Collinear (parallel) vectors

$\vec{u} = (a ; b)$ and $\vec{v} = (c ; d)$ are collinear iff $a d - b c = 0$.

This is the determinant: cross product (main diagonal − secondary diagonal).

Vector norm (length)

$\vec{u} = (a ; b)$:

$\|\vec{u}\| = \sqrt{a^2 + b^2}$

Pythagoras applied to coordinates.

Common mistakes

1. Swapping $B$ and $A$ in “arrival minus departure”.
2. Confusing coordinates of $\overrightarrow{AB}$ with those of $A$.
3. Writing midpoint = $(x_B - x_A) / 2$ (it’s $+$ for midpoint, $-$ for a vector).

?À toi de jouer

A(1 ; 2) and B(4 ; 8). Coordinates of AB?

A.(3 ; 6)B.(4 ; 8)C.(−3 ; −6)D.(5 ; 10)

Valider