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.

Slope

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

Hence:

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.

Eddie

This blog is property of Edward Shore, 2017.
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