Find the Intersection of Two Line Segments in 2D (Easy Method)

Рет қаралды 3,765

11 ай бұрын

In this video, I will show you how to find the intersection of two line segments in 2D. This is a simple but important concept in geometry, and it can be used in many different applications. I will explain the method step-by-step, and I will also provide some examples.
Points:
A(x1,y1)
B(x2,y2)
C(x3,y3)
D(x4,y4)
Given two line segments AB and CD, find the point of intersection P(x0, y0).
Solution:
Vectors:
A = (x1, y1)
B = (x2, y2)
C = (x3, y3)
D = (x4, y4)
AB = (x2 - x1, y2 - y1)
CD = (x4 - x3, y4 - y3)
A + αAB = C + βCD
α = ((x4 - x3)(y3 - y1) - (y4 - y3)(x3 - x1)) / ((x4 - x3)(y2 - y1) - (y4 - y3)(x2 - x1)) = a / b
β = ((x2 - x1)(y3 - y1) - (y2 - y1)(x3 - x1)) / ((x4 - x3)(y2 - y1) - (y4 - y3)(x2 - x1)) = c / b
P = (x0, y0)
x0 = x1 + α(x2 - x1) = x3 + β(x4 - x3)
y0 = y1 + α(y2 - y1) = y3 + β(y4 - y3)
/ edgardocpu

Пікірлер: 6

@glebickapimpin9852 7 ай бұрын

Thank you for simple and fast explanation.

@GoalWalker 9 ай бұрын

Next video topic idea: If two line segments are collinear, how to find the overlap? Like defining the overlap as its own line segment and finding each endpoint.

@GoalWalker 9 ай бұрын

2:30 If a=b=0 means the line segments are collinear, does c=b=0 also mean the line segments are collinear?

@GoalWalker 9 ай бұрын

1:50 Can you go into more steps to work out equations for alpha? Each time I do, I get different order for numerator for alpha, like if the two multiplying parts were switched. As well as, I have the denominator for alpha and beta are different. I don't know where I am going wrong: (the workout below is from my handwork, the captions for each step were rewritten from chatGPT) alpha*(x2-x1) - beta*(x4-x3) = x3-x1 alpha*(y2-y1) - beta*(y4-y3) = y3-y1 Solve for alpha: # Solve for alpha in terms of beta: beta = (alpha*(x2-x1) - (x3-x1)) / (x4-x3) beta = (alpha*(y2-y1) - (y3-y1)) / (y4-y3) # Since both equations equal alpha, they must be equal to each other: [alpha*(x2-x1) - (x3-x1)] / (x4-x3) = [alpha*(y2-y1) - (y3-y1)] / (y4-y3) # Cross-multiply to eliminate fractions: (y4-y3)*(alpha*(x2-x1)-(x3-x1)) = (alpha*(y2-y1)-(y3-y1))*(x4-x3) # Expand and rearrange the equation to isolate alpha: (alpha*(x2-x1)*(y4-y3)) - (x3-x1)*(y4-y3) = (alpha*(x4-x3)*(y2-y1)) - (x4-x3)*(y3-y1) # Now, isolate alpha: (alpha*(x2-x1)*(y4-y3)) - (alpha*(x4-x3)*(y2-y1)) = (x3-x1)*(y4-y3) - (x4-x3)*(y3-y1) # Factor out alpha: alpha*[(x2-x1)*(y4-y3) - (x4-x3)*(y2-y1)] = (x3-x1)*(y4-y3) - (x4-x3)*(y3-y1) # Finally, solve for alpha: alpha = [(x3-x1)*(y4-y3) - (x4-x3)*(y3-y1)] / [(x2-x1)*(y4-y3) - (x4-x3)*(y2-y1)] As you can see from alpha, the two products in the numerator and denominator are switched from the alpha formula in the video. Why? Note 1: When I ask chatGPT, they just say my result is correct. Which to me seems wrong. Note 2: The equations for alpha and beta in video appear in a different order than to t and u from the wiki of "Given two points on each line segment"

@GoalWalker 9 ай бұрын

I got alpha as in the video by isolating alpha to the right side, rather than the left. But I got beta as in the video by isolating to the left. Weird, especially when doing by hand.

@jasonmakepeaceranchohs869 3 ай бұрын

I got his results by setting it up as a matrix equation and using Cramer's Rule. I subtracted x1 and y1 to the right and β(x4 - x3) and β(y4 - y3) to the left. Cramer's Rule for α and β worked pretty slick. Might have to adjust some of the commutative order on a few things to make it look the same, but it worked out for me.