Saturday, August 12, 2017

Geometry: The Intersection Point of a Quadrilateral

Geometry: The Intersection Point of a Quadrilateral

The Setup

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)

The Intercept

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.

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