# Intersection of two lines

I'm using the following equation to represent lines

(x, y) = (start_x, start_y) + t((end_x, end_y) - (start_x, start_y))

I'm trying to find the interesection point of two lines written in this form.

I have been able to solve for t and plug it back into the equation, but i get two values of t when i solve for it. So that gives me 2 possible x values and 2 possible y values in the end. One x value is correct, and one y value is correct, but I get a wrong x value and a wrong y value.

I'm wondering if theres a totally different way, or a way to get rid of the x and y value that I don't need.

Thanks.

arildno
Homework Helper
Gold Member
Dearly Missed
Welcome to PF!
The parameter "t" need not have the same value at the point of intersection for each line!!
What you need, is to solve a 2*2 system of two unknowns, the two parametrization variables:
Let superscript "1" denote Line 1 expression, "2" Line 2 expression.
Then we may write:
$$(x,y)=(x^{1}_{0},y^{1}_{0})+t^{1}(x_{1}^{1}-x_{0}^{1}, y_{1}^{1}-y_{0}^{1})$$
And line 2:
$$(x,y)=(x^{2}_{0},y^{2}_{0})+t^{2}(x_{1}^{2}-x_{0}^{2}, y_{1}^{2}-y_{0}^{2})$$
Set these expressions equal to each other, and solve for $$t^{1},t^{2}$$

You may also eliminate the parametrization variable, and solve the following system
for x and y:
$$y=a^{1}x+b^{1}$$
$$y=a^{2}x+b^{2}$$

Ok, thanks.

What I ended up doing was turning the (x, y) = (a, b) + t(c - a, d - b) into y = mx + b. Then I solved for x and y.

The reason I wanted t was to know whether the intersection was actually between the two points (0 <= t <= 1). But I can just plug the point I found back into the first equation and solve for t.

x = a + t(c - a)
t = (x - a) / (c - a)