Using Coordinate Geometry to Prove that a Quadrilateral is a Parallelogram

Question: How can we determine whether a quadrilateral placed on a coordinate plane is a parallelogram of not?
There are a number of ways to show whether a quadrilateral placed on a coordinate plane is a parallelogram or not. Here are a few ways:[br][br]1. Show that both pairs of opposite sides are congruent. [br]2. Show that both pairs of opposite sides are parallel[br]3. Show that a pair of opposite sides are congruent and parallel[br]4. Show that the diagonals bisect each other.[br][br]In this activity, we will use the Distance, Midpoint and Slope Formulas that we learned in Algebra 1 to show congruent, bisected and parallel segments.
Midpoint, Slope and Distance Formula Review.
The midpoint of a segment in the coordinate plane with endpoints [math]A(x_1,y_1)[/math] and [math]B(x_2,y_2)[/math] is given by[br][br] [math]M_{AB}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)[/math][br][br]For example, the midpoint of segment [math]AB[/math] with endpoints [math]A\left(-3,7\right)[/math] and [math]B\left(7,5\right)[/math] is:[br][br] [math]M_{AB}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{-3+7}{2},\frac{7+5}{2}\right)=\left(2,6\right)[/math][br][br]The slope ([math]m[/math]) of a line on a coordinate plane is found using the formula[br][br] [math]m=\frac{change-in-y}{change-in-x}=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}[/math][br][br]For example, the slope of the line that passes through [math]A\left(-3,7\right)[/math] and [math]B\left(7,5\right)[/math] is:[br][br] [math]m=\frac{y_2-y_1}{x_2-x_1}=\frac{5-7}{7-\left(-3\right)}=\frac{-2}{10}=-\frac{1}{5}[/math][br][br]The distance between points [math]A\left(-3,7\right)[/math] and [math]B\left(7,5\right)[/math] (i.e. the length of segment [math]AB[/math]) is found using the distance formula:[br] [br] [math]AB=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}[/math][br][br]For the points shown,[br] [math]AB=\sqrt{\left(7-\left(-3\right)\right)^2+\left(5-7\right)^2}=\sqrt{10^2+\left(-2\right)^2}=\sqrt{104}=10.2[/math]
Exploring the Distance Formula
The distance formula given above can be written as: [math]\left(AB\right)^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2[/math][br]This is precisely the Pythagorean Theorem [math]c^2=a^2+b^2[/math][br] if we make the substitutions:[br] [math]c=AB[/math], [math]a=\left(x_2-x_1\right)[/math] and [math]b=\left(y_2-y_1\right)[/math].[br][br]In the applet below, a quadrilateral has been drawn on a coordinate plane. Using the sides as hypotenuses, we construct right triangles on the four sides of the quadrilateral. Two of the right triangles have been drawn. draw the other two and complete the spreadsheet at the right. To enter the measured length of a segment into the spreadsheet, just type the endpoints of the segment. Then use the distance formula to calculate the length of the segment. Compare the measured and calculated lengths.
Is the quadrilateral a parallelogram?
Based on your side length measurements and calculations can you conclude that the quadrilateral is a parallelogram? Give reason(s) why or why not.
Prove that a quadrilateral is a parallelogram
Instead of measuring and/or calculating the side lengths, we would like to prove that the opposite sides of the quadrilateral are congruent using the right triangles we constructed. Which of the following postulates or theorems could we use to prove the right triangles congruent based on the information in our sketch?
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Information: Using Coordinate Geometry to Prove that a Quadrilateral is a Parallelogram