**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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