Question

The coordinates of the vertices of rhombus ABCD are A(1, 1), B(5, 3), C(7, 7), and D(3, 5). Find the coordinates of the point of intersection of the diagonals.

Answer 1

give me twenty seconds to make a digital copy of this shape

Answer 2

and show your WORK

Answer 3

@karama i appreciate your kindness

Answer 4

you're very welcome :)

Answer 5

ohh, man. Thanks for taken your time

Answer 6

Alright! Let's give our question a shot. Our rhombus ABCD has vertices A(1, 1), B(5, 3), C(7, 7), and D(3, 5); using the properties of a rhombus, we also know that it will have four sides of equal length (a). We also have the knowledge ( http://www.algebra.com/algebra/homework/word/geometry/The-length-of-diagonals-of-a-rhombus.lesson) that the diagonal of a rhombus is equal to\[ l= 2a*\cos(\theta/2)\]
I know you wanted my work on the formula, but I'm not up to doing a fresh derivative at the moment; however, the link provides a proof.

Using our givens, we can plug in our side length a
with is equal to
\[a=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1}^{2})}\]
where \[x_{i}\] and \[y_{i}\] are the x and y values of any two points not forming our diagonals.
I got the side formula from length of a diagonal.
by using A and B
\[a=\sqrt{(16+4)}\]
\[l=2*\sqrt{20}*cos(\theta/2)\] I can't remember how to get the angles; after you do get them, however, you can draw to diagonals AC and BD, draw a point at their intersection their length. The coordinates of that point will be your answer.

Answer 7

@Powerful22 I'm pretty sure you can get the angles using a pythagorean identity. Make triangles out of your sides and diagonals then solve for the angle.