Thread: Rhombus Coordinate Geometry

Find the coordinates of C and D, such that ABCD is a rhombus. (see attached diagram)

I went ahead with simultaneous derived from distance formulas and perpendiular gradients and then solved that with Pythagoras theorem to end up with the points
C(10.5, 13.5) and D(5.5, 1.5).

Feeling very pleased with myself for solving it, i realised i had made an assumption that was incorrect. I assumed that ABCD was a square, and using Pythagoras, with AB = 13 units, then side AD = and proceeded from there and it all worked fine.

Realising now the assumption was faulty, i have tried to do it without the property of ABCD being a square, and just cannot make any headway with it.

Any help greatly appreciated

Well, the question lacks information, because there is an infinite number of solutions, and remember that a square is a hombus, and hence, your solution is one of the many which exists.

The line through AB has equation 12y = -5x + 130

The midpoint AB is (8, 7.5)

The line perpendicular to AB through (8, 7.5) has equation: 10y = 24x - 117

So basically, any pair of points equidistant from (12, 7.5) and on the line 10y = 24x - 117 is a solution.

Since your points satisfy this, they are correct

Unless more information are given.

Thanks Unknown008, that was the conclusion that I came to as well, after hours of trying every system of equations i could think of, that perhaps it was meant to be a square rather than a rhombus, or there was something lacking in the question. Just thought i would ask here in case I am missing something and there is a way to do it becuase it is bugging me now lol

I was pondering the square is a rhombus as a way to justify my assumption. But while are squares are rhombuses, not all rhombuses are squares which means technically my assumption is faulty, but I just cannot see any other way around it otherwise.