# Vertices of a parallelogram?

Printable View

• Mar 20th 2008, 01:04 AM
oXCryssieLeahXo
Vertices of a parallelogram?
Goodness I don't understand this at all!!! I understand finding the coordinates of a parallelogram, but this makes no sense.

So it says "Which of the set of points cannot be the vertices of parallelogram ABCD?"

options are

a) A (-4, -2), B (2, 0), C (3, 3), D (-2, -1)

b) A (-2, -1), B (3, -2), C (4, 1), D (-1, 2)

c) A (0, -2), B (4, -1), C (5, 2), D (0, 5)

d) A (0, -2), B (5, -3), C (7, 1), D (2, 2)

so can someone tell me what formula to use to find the answer? (I'm assuming there's a formula!)
• Mar 20th 2008, 02:02 AM
earboth
Quote:

Originally Posted by oXCryssieLeahXo
...

So it says "Which of the set of points cannot be the vertices of parallelogram ABCD?"

options are

a) A (-4, -2), B (2, 0), C (3, 3), D (-2, -1)

b) A (-2, -1), B (3, -2), C (4, 1), D (-1, 2)

c) A (0, -2), B (4, -1), C (5, 2), D (0, 5)

d) A (0, -2), B (5, -3), C (7, 1), D (2, 2)

...

The properties of a parallelogram are:

1. The diagonals have a common midpoint.
2. 2 pairs of parallel sides
3. The parallel sides have the same length (#3 is equivalent to #2)

First check if
$\displaystyle M_{AC} = M_{BD}$

So only b) and d) could be a parallelogram

Check if

$\displaystyle (\overline{AB})\ \parallel \ (\overline{CD})$

only d) is left. Since $\displaystyle \left(\begin{array}{c}-5\\1\end{array}\right) = k \cdot \left(\begin{array}{c}5\\-1\end{array}\right)$ the sides $\displaystyle (\overline{AB})\ \parallel \ (\overline{CD})$

PS: Fröhliche Ostern!
• Mar 20th 2008, 02:15 AM
oXCryssieLeahXo
Thank yah
Quote:

Originally Posted by earboth
The properties of a parallelogram are:

1. The diagonals have a common midpoint.
2. 2 pairs of parallel sides
3. The parallel sides have the same length (#3 is equivalent to #2)

First check if
$\displaystyle M_{AC} = M_{BD}$

So only b) and d) could be a parallelogram

Check if

$\displaystyle (\overline{AB})\ \parallel \ (\overline{CD})$

only d) is left. Since $\displaystyle \left(\begin{array}{c}-5\\1\end{array}\right) = k \cdot \left(\begin{array}{c}5\\-1\end{array}\right)$ the sides $\displaystyle (\overline{AB})\ \parallel \ (\overline{CD})$

PS: Fröhliche Ostern!

wow thank you SO much. Makes sense. I kinda feel dumb now that I didn't figure that out myself!! (Headbang)

Frohe Ostern an dir auchh! =]]
• Mar 20th 2008, 05:12 AM
Soroban
Hello, oXCryssieLeahXo!

I don't suppose you tried to PLOT the points . . .

Quote:

Which of the set of points cannot be the vertices of parallelogram ABCD?

. . $\displaystyle a)\;A (\text{-}4, \text{-}2),\;B (2, 0),\;C (3, 3), D\;(\text{-}2, \text{-}1)$

Code:

                            |          C         .  .  .  .  .  +  .  .  o  .                             |         .  .  .  .  .  +  .  .  .  .                             |         .  .  .  .  .  +  .  .  .  .                             |       - + - + - + - + - + - + - + - o - + - + -                     D      |      B         .  .  .  o  .  +  .  .  .  .                             |         .  o  .  .  .  +  .  .  .  .             A              |
No, I don't think so . . .