How would you do coordinate geometry proofs?

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How would you do coordinate geometry proofs?

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Answer 1 · 2016-03-01T04:10:50

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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 $A (x_{A}, y_{A})$ $A(x_A,y_A)$ , $B (x_{B}, y_{B})$ $B(x_B,y_B)$ , $C (x_{C}, y_{C})$ $C(x_C,y_C)$ and $D (x_{D}, y_{D})$ $D(x_D,y_D)$ are vertices of any quadrilateral with coordinates given in parenthesis.

Midpoint $P$ $P$ of $A B$ $AB$ has coordinates
$(x_{P} = \frac{x_{A} + x_{B}}{2}, y_{P} = \frac{y_{A} + y_{B}}{2})$ $(x_P=(x_A+x_B)/2,y_P=(y_A+y_B)/2)$
Midpoint $Q$ $Q$ of $A D$ $AD$ has coordinates
$(x_{Q} = \frac{x_{A} + x_{D}}{2}, y_{Q} = \frac{y_{A} + y_{D}}{2})$ $(x_Q=(x_A+x_D)/2,y_Q=(y_A+y_D)/2)$
Midpoint $R$ $R$ of $C B$ $CB$ has coordinates
$(x_{R} = \frac{x_{C} + x_{B}}{2}, y_{R} = \frac{y_{C} + y_{B}}{2})$ $(x_R=(x_C+x_B)/2,y_R=(y_C+y_B)/2)$
Midpoint $S$ $S$ of $C D$ $CD$ has coordinates
$(x_{S} = \frac{x_{C} + x_{D}}{2}, y_{S} = \frac{y_{C} + y_{D}}{2})$ $(x_S=(x_C+x_D)/2,y_S=(y_C+y_D)/2)$

Let's prove that $P Q$ $PQ$ is parallel to $R S$ $RS$ . For this, let's calculate the slope of both and compare them.

$P Q$ $PQ$ has a slope
$\frac{y_{Q} - y_{P}}{x_{Q} - x_{P}} = \frac{y_{A} + y_{D} - y_{A} - y_{B}}{x_{A} + x_{D} - x_{A} - x_{B}} =$ $(y_Q-y_P)/(x_Q-x_P)=(y_A+y_D-y_A-y_B)/(x_A+x_D-x_A-x_B)=$
$= \frac{y_{D} - y_{B}}{x_{D} - x_{B}}$ $=(y_D-y_B)/(x_D-x_B)$

$R S$ $RS$ has a slope
$\frac{y_{S} - y_{R}}{x_{S} - x_{R}} = \frac{y_{C} + y_{D} - y_{C} - y_{B}}{x_{C} + x_{D} - x_{C} - x_{B}} =$ $(y_S-y_R)/(x_S-x_R)=(y_C+y_D-y_C-y_B)/(x_C+x_D-x_C-x_B)=$
$= \frac{y_{D} - y_{B}}{x_{D} - x_{B}}$ $=(y_D-y_B)/(x_D-x_B)$

As we see, the slopes of $P Q$ $PQ$ and $R S$ $RS$ are the same.
Analogously, slopes of $P R$ $PR$ and $Q S$ $QS$ are the same as well.

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

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How would you do coordinate geometry proofs?

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