Program for Point of Intersection of Two Lines

Given points A and B corresponding to line AB and points P and Q corresponding to line PQ, find the point of intersection of these lines. The points are given in 2D Plane with their X and Y Coordinates. Examples:

Input : A = (1, 1), B = (4, 4)
 C = (1, 8), D = (2, 4)
Output : The intersection of the given lines 
 AB and CD is: (2.4, 2.4)

Input : A = (0, 1), B = (0, 4)
 C = (1, 8), D = (1, 4)
Output : The given lines AB and CD are parallel.

First of all, let us assume that we have two points (x₁, y₁) and (x₂, y₂). Now, we find the equation of line formed by these points. Let the given lines be :

1. a₁x + b₁y = c₁
2. a₂x + b₂y = c₂

We have to now solve these 2 equations to find the point of intersection. To solve, we multiply 1. by b₂ and 2 by b₁ This gives us, a₁b₂x + b₁b₂y = c₁b₂ a₂b₁x + b₂b₁y = c₂b₁ Subtracting these we get, (a₁b₂ - a₂b₁) x = c₁b₂ - c₂b₁ This gives us the value of x. Similarly, we can find the value of y. (x, y) gives us the point of intersection. Note: This gives the point of intersection of two lines, but if we are given line segments instead of lines, we have to also recheck that the point so computed actually lies on both the line segments. If the line segment is specified by points (x₁, y₁) and (x₂, y₂), then to check if (x, y) is on the segment we have to just check that

• min (x₁, x₂) <= x <= max (x₁, x₂)
• min (y₁, y₂) <= y <= max (y₁, y₂)

The pseudo code for the above implementation:

determinant = a1 b2 - a2 b1
if (determinant == 0)
{
 // Lines are parallel
}
else
{
 x = (c1b2 - c2b1)/determinant
 y = (a1c2 - a2c1)/determinant
}

These can be derived by first getting the slope directly and then finding the intercept of the line. 

Output:

The intersection of the given lines AB and 
CD is: (2.4, 2.4)

Time Complexity: O(1) 
Auxiliary Space: O(1)