Area of a rhombus in co-ordinate geometry

Question

I am trying to solve the question:

In the $xy$-plane, two adjacent vertices of the rhombus are $(-2,2)$ and $(2,7)$

A. Area of rhombus is greater

B. 41 is greater

C. Both are equal

D. Cannot be determined

I tried the distance formula to get the distance between the two points which is $\sqrt{41}$ but I don’t know what to do next.

Answer 1

The area cannot be determined as two points are insufficient to describe a rhombus.

Answer 2

Using https://artofproblemsolving.com/wiki/index.php/Rhombus#Proof_that_the_diagonals_of_a_rhombus_divide_it_into_4_congruent_triangles

the intersection of the two diagonals will lie on the circle

$$(x-2)(x+2)+(y-2)(y-7)=0$$

Hence the point cannot be determined with the information supplied

So, will be the lengths of the two diagonals

Answer 3

Since the side of the rhombus is $\sqrt{41}$, the area of the rhombus is $41\sin\theta$, where $\theta$ is either of the (supplementary) internal angles of the rhombus. So the area is somewhere between $0$ (if the rhombus is totally "flat") and $41$ (if the rhombus is a square). If we were told that the rhombus is not a square, B would be the correct answer. As we were not told that, D is the correct answer.