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.
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)
Hello, euclid2!
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.