Thread: co-ordinates of a parallelogram

1. co-ordinates of a parallelogram

ABCD is a parallelogram where A is (4,2), B is (-6,1), and D is (-3,-4). Find the co-ordinates of C.

2. If you have a sketch, that should be simpler.

The properties of a parallelogram is such that the sides AB and CD are parallel, meaning that AB and CD have the same gradient.
Same for BC and AD; they have the same gradient.

So:
1. Find the gradient of AB. (this is also gradient of CD)
2. Find the gradient of AD. (this is also gradient of BC)
3. Find equation of line through D having gradient AB.
4. Find equation of line through B having gradient AD.
5. Solve simultaneously for those two lines. The intersection is where point C is.

There is a shorter, visual way using vectors, but this makes a diagram a must, if you cannot picture the points in your head.

Vector AB = (10, 1)
So, C + (10, 1) = (-3, -4)
So, C = (-3 -10, -4 -1) = (-13, -5)

3. Hello, euclid2!

If you make a sketch, you can "walk" your way to the answer.

$ABCD\text{ is a parallelogram with vertices: }A(4,2),\;B(\text{-}6,1),\;D(\text{-}3,\text{-}4)$
$\text{Find the coordinates of }C.$
Code:
                              A
B               o(4,2)
(-6,1)o               :
:               :
:               : -6
:               :
:       D   -7  :
C       :       o - - - +
o - - - +    (-3,-4)

We see that vertex $\,C$ is at the lower-left.

Going from $\,A$ to $\,D$, we move: 6 units down and 7 units left.

Since $BC \parallel AD$, going from $\,B$ to $\,C$, we do the same.

Starting at $B(\text{-}6,1)$, move 6 units down a 7 units left.

Therefore, vertex $\,C$ is $(\text{-}13,\text{-}5)$

4. Originally Posted by euclid2
ABCD is a parallelogram where A is (4,2), B is (-6,1), and D is (-3,-4). Find the co-ordinates of C.
B(-6 , 1) , A(4 , 2)
C(a , b) , D(-3 , -4)

4 - (-6) = -3 - a
10 = -3 - a ===> a = -13

2 - 1 = -4 - b
1 = -4 -b ===> b = -5

therefore C(-13 , -5)

5. Re: co-ordinates of a parallelogram

is it just me or is -13, -5 making the parallelogram's C point jut far out compared to the other points when graphing, that can't be right. 7, -3 sounds more accurate but I'm also having trouble figuring out how to arrive at that answer without just testing it on a graph.

6. Re: co-ordinates of a parallelogram

Figured it out,

You need to solve for AB = CD. AB you previously calculated its distance was equal to (-10,-1). so you set x-10 and y-1 and plug in the D point (-3,-4). Now you have x-10=-3 and y-1=-4. Once you've calculated that you'll find C = (7,-3). If you want to confirm this, go ahead and graph the points, A,B,C,D on GeoGebra and you'll see it makes a perfect parallelogram.

7. Re: co-ordinates of a parallelogram

Well geezzz, much easier if you change D to C !!

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