Find the ending coordinates of line if it is extended by a D distance

Question

A line draw between $(x_1,y_1)$ to $(x_2,y_2)$. I need to find the ending point in which the line is extended by $D$ distance(or length).

From the above image, $(x_1,y_1)$,$(x_2,y_2)$ and $D$ is known, I need to find $(x_3,y_3)$.

Answer 1

Welcome to MSE. One way to express the location of $(x_3, y_3)$ is as $\langle x_2, y_2 \rangle + D \cdot \langle u, v \rangle $, where $\langle u, v\rangle$ is the unit vector pointing in the direction of $\langle x_2, y_2 \rangle - \langle x_1, y_1 \rangle$

Answer 2

Define the unit vector from $(x_1, y_1)$ to $(x_2, y_2)$ as follows

$ v = \dfrac{ (x_2 - x_1, y_2 - y_1) }{L}$

where $L = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } $

Then

$(x_3, y_3) = (x_1, y_1) + (L+D) v $

or equivalently,

$(x_3, y_3) = (x_2, y_2) + D v $

Answer 3

The line through $(x_1, y_1)$ and $(x_1, y_2)$ can be written $y= \frac{y_2-y_1}{x_2-x_2}(x- x_2)+ y_2$.

The circle with center at $(x_2, y_2)$ and radius D can be written $(x-x_2)^2+ (y-y_2)^2= D^2$.

$(x_3, y_3)$, the endpoint of the line segment extended from $(x_2, y_2)$ by distance D is where that line and circle intersect so we must have $(x_3- x_2)^2+ (\frac{y_2-y_1}{x_2-x_2}(x_3- x_2))^2= D^2$.

Solve that quadratic equation for $x_3$ and then use the equation of the line to find $y_3$.

(That equation will have two roots. You want the one on the other side of $x_2$ from $x_1$.)