Question 1040566: prove by three methods that the points (5,3), (-8,-5), (-6,-6) and (7,2) form a parallelogram??
Found 2 solutions by Edwin McCravy, Boreal:
Answer by Edwin McCravy(20081)   (Show Source):
You can put this solution on YOUR website! 
There are four methods to show it:
1. A parallelogram is a quadrilateral with both pairs of
opposite sides parallel.
A. Use the slope formula to show that AB is parallel to CD
B. Use the slope formula to show that BC is parallel to AD
2, If both pairs of opposite sides of a quadrilateral are
congruent, the quadrilateral is a parallelogram.
A. Use the distance formula to show that AB and CD have the
same length and therefore are congruent.
B. Use the distance formula to show that BC and AD have the
same length and therefore are congruent.
3. If the diagonals of a quadrilateral bisect each other,
the quadrilateral is a parallelogram.
A. Use the midpoint formula to find the midpoint of diagonal AC.
B. Use the midpoint formula to find the midpoint of diagonal BD
to show that the two midpoints are the same.
4. If ONE PAIR of opposite sides of a quadrilateral are BOTH
parallel and congruent, the quadrilateral is a parallelogram.
If you have done both 1 and 2, you have already done enough to
show that AD and BC are both parallel and congruent.
If you have trouble, tell me in the thank-you note form below and
I'll get back to you by email.
Edwin
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! Draw it. The lines between the two Quadrant III points and the other two points should have the same slope and be parallel.
The slope is -1/2 between (-8,-5) and (-6,-6). It is -1/2 for the other two points, too.
Between (-6,-6) and (7.2) slope is 8/13. That should be the same between (5,3) and (-8.-5). It is.
The sides should be equal length. The distance between the third quadrant points is sqrt(1^2+2^2) The distance between the two first quadrant points is (1^2+2^2). For the longer sides, (-6,-6) and (7,2), it is sqrt (169+64), and for the other two (5,3) and (-8,-5), it is sqrt (64+169).
For the angles, one may show that the diagonals intersect at their midpoint. Then the triangles formed would be congruent on the basis of SAS (the angles being vertical angles).
Diagonal from (-6,-6) to (5,3) has slope 9/11 and its equation is y-3=(9/11)(x-5)
That is y=(9/11)x-(12/11)
The other diagonal is from (-8,-5) to (7,2). Its slope is (7/15) and function is y-2=(7/15)(x-7). That is
y=(7/15)x-19/15
The intersection of the two diagonals is solving this system.
11y-9x=-12
15y-7x=-19
77y-63x=-84
-135y+63x=+171
-58y=87; y=-3/2
-16.5-9x=-12
-9x=4.5; x=-0.5
Check into second. -22.5+3.5=-19
intersect at (-0.5,-1.5)
The midpoint of the first diagonal is (-.5,-1.5)
The midpoint of the second diagonal is (-.5,-1.5)
The angles between the half diagonals are equal.
SAS and the angles are equal.
|
|
|
