ABCD is a parallelogram where A is (4,2), B is (-6,1), and D is (-3,-4). Find the co-ordinates of C.
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.
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)
If you make a sketch, you can "walk" your way to the answer.
Code:A B o(4,2) (-6,1)o : : : : : -6 : : : D -7 : C : o - - - + o - - - + (-3,-4)
We see that vertex is at the lower-left.
Going from to , we move: 6 units down and 7 units left.
Since , going from to , we do the same.
Starting at , move 6 units down a 7 units left.
Therefore, vertex is
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.
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.