How would you do coordinate geometry proofs?

1 Answer
Mar 1, 2016

See below

Explanation:

Coordinate proof is an algebraic proof of a geometric theorem. In other words, we use numbers (coordinates) instead of points and lines.

In some cases to prove a theorem algebraically, using coordinates, is easier than to come up with logical proof using theorems of geometry.

For example, let's prove using the coordinate method the Midline Theorem that states:
Midpoints of sides of any quadrilateral form a parallelogram.

Let four points #A(x_A,y_A)#, #B(x_B,y_B)#, #C(x_C,y_C)# and #D(x_D,y_D)# are vertices of any quadrilateral with coordinates given in parenthesis.

Midpoint #P# of #AB# has coordinates
#(x_P=(x_A+x_B)/2,y_P=(y_A+y_B)/2)#
Midpoint #Q# of #AD# has coordinates
#(x_Q=(x_A+x_D)/2,y_Q=(y_A+y_D)/2)#
Midpoint #R# of #CB# has coordinates
#(x_R=(x_C+x_B)/2,y_R=(y_C+y_B)/2)#
Midpoint #S# of #CD# has coordinates
#(x_S=(x_C+x_D)/2,y_S=(y_C+y_D)/2)#

Let's prove that #PQ# is parallel to #RS#. For this, let's calculate the slope of both and compare them.

#PQ# has a slope
#(y_Q-y_P)/(x_Q-x_P)=(y_A+y_D-y_A-y_B)/(x_A+x_D-x_A-x_B)=#
#=(y_D-y_B)/(x_D-x_B)#

#RS# has a slope
#(y_S-y_R)/(x_S-x_R)=(y_C+y_D-y_C-y_B)/(x_C+x_D-x_C-x_B)=#
#=(y_D-y_B)/(x_D-x_B)#

As we see, the slopes of #PQ# and #RS# are the same.
Analogously, slopes of #PR# and #QS# are the same as well.

So, we have proven that opposite sides of quadrilateral #PQRS# are parallel to each other. That is a sufficient condition for this object to be a parallelogram.