ABCD is a parallelogram where A is (4,2), B is (-6,1) and D is (-3, -4). Find the coordinates of C.
I can find the length of AD, of AB and of BD but I don't know where to go from there.
I calculated the lengths of 2 of the sides as the square root of 101 and the square root of 34. I can say that DC is the square root of 101 and AC is the square root of 34 but where does that place C?
What formula are you using, that you need to start by finding the lengths of the sides? :shock:
Are you allowed to use the regular formula? If so, apply it to the vectors in question. If not, then please explain the method that you are instead required to use.
ABCD is a parallelogram where A is (4,2), B is (-6,1) and D is (-3, -4). Find the coordinates of C.
I can find the length of AD, of AB and of BD but I don't know where to go from there.
Thank you. I just figured it out. I just have to get the lengths between B and A and between B and D and apply these lengths in terms of x and y to either A or D as a reference point and I get C's co-ordinates to be (7, -3)
Assuming C is in the third quadrant, then vector AB = vector DC and let point C = (x,y).
vector AB = <-10,-1> and vector DC = <x+3,y+4>
Hence x+3 = -10, x = -13 and y+4 = -1, y = -5, ergo C = (x,y) = (-13,-5).
Note: A bit of geometry. If you are given three non-colinear points in a plane and you must find the fourth point to make a parallelogram, you have three options on where to place the fourth point.
How could C be in the third quadrant since A is in the 1st quadrant, B is in the 2nd quadrant and D is in the 3rd quadrant. Sketching it roughly on graph paper, I get C in the 4th quadrant.
In fact Mickey (to be mathematically correct) the coordinate(s) of C (fourth point) are (1,7) or (7,-3) or (-13,-5) will give a parallelogram with the other three points.
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