Geometry: The Intersection Point of a Quadrilateral
We are given four points, A, B, C, and D, as four Cartesian coordinates. We connect the four points, starting with A, in a clockwise form to form a quadrilateral. We will designate each points as the coordinates:
A: (ax, ay)
B: (bx, by)
C: (cx, cy)
D: (dx, dy)
Draw a line from one corner to the opposite corner. This results in two lines: AC and DB. The two lines (show above in lime green) meet at point P. The goal is determine the coordinates of P.
We know the equation of the line is y = m*x + b, where m is the slope and b is the y-intercept. Note that:
y = m*x + b
y – m*x = b
In geometry, the slope of a line containing two points is generally defined as:
m = (change in y coordinates)/(change in x coordinates) = Δy/Δx = (y2 – y1)/(x2 – x1)
The slope of AC: SAC = (cy – ay)/(cx – ax)
The slope of BD: SBD = (dy – by)/(dx –bx)
We can easily deduce that solving the general equation of the line y = m*x + b for the intercept yields b = y – m*x. If follows that:
The intercept of AC: IAC = cy – SAC * cx = ay – SAC *ax
The intercept of BD: IBD = by – SBD * bx = dy – IBD *dx
Finding the Intersection Point
Now that the slopes and intercepts are determined, we can form the following system of equations:
(I) y = SAC * x + IAC
(II) y = SBD * x + IBD
Solving for x and y will find our intersection point P (px, py). Subtracting (II) from (I) (see above):
0 = (SAC – SBD) * x + (IAC – IBD)
We can solve for x.
-(IAC – IBD) = (SAC – SBD) * x
(-1*IAC - -1*IBD) = (SAC – SBD) * x
(-IAC + IBD) = (SAC – SBD) * x
(IBD – IAC) = (SAC – SBD) * x
x = px = (IBD – IAC)/(SAC – SBD)
It follows that we determine y by either equation (I) or (II):
y = py = SAC * px + IAC = SBD * px + IBD
Our desired coordinates of point P are found.
This blog is property of Edward Shore, 2017.