Calculate a point at a distance perpendicular to a vector offset (2D)

To calculate a point at a distance perpendicular to a vector offset

Point perpendicular to offset

First the point on the vector $P'$ at the required offset needs to be calculated. This easy to do:

$P' = A + (B-A) * offset$

$P'_x = A_x + (B_x - A_x) * offset$
$P'_y = A_y + (B_y - A_y) * offset$

The vector to move the point P’ on is perpendicular to $\overline{AB}$ . This vector can be calculated using a 90 degree rotation matrix:

The matrix shows in the the first row $x'= -y$ and the second row shows $y'= x$

You can safely ignore the stuff above if you are only interested in getting the job done. The final formula is:

$\begin{pmatrix}x'\\y' \end{pmatrix}= \begin{pmatrix}-y\\x \end{pmatrix}$

We will call the vector ( $\overline{P'P}$ ) vector $C$ .

$C =\begin{pmatrix}- (B_y - A_y)\\ B_x - A_x \end{pmatrix}$

Vector $C$ needs to be normalized (divided by it’s length / magnitude).

$\hat{C} =\dfrac{C}{\left \|C \right \|}$

The resulting formula becomes:

$P = (A + (B-A) * offset) + \hat {C} * distance$

A positive distance will get the point above the vector, a negative below the vector.

Developer notes