Show that the four points given below are the vertices of a rhombus.

Question

Show that the four points, $(5, 8), (7, 5), (3, 5)$ and $(5, 2)$ are the vertices of a rhombus.

I tried solving it, by finding out the distances by using the formula $\sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$

I found all of them but two sides are equal and other two are not. As far as I know, rhombus has all sides equal...but mine does not match! Help please :)

Answer 1

Hint: Here are the points on graph paper.

Answer 2

The sides are formed by joining the points as follows: $(5,8) - (7,5)$, $(7,5) - (5,2)$, $(5,2) - (3,5)$, and $(3,5)-(5,8)$. If you check these four lengths, you will find that they are all equal.

Answer 3

Since we do not know in which order the above mentioned vertices come, 6 different line segments can be formed, (4 edges and 2 diagonals in case of a rhombus)

If the shape is indeed a rhombus, then if we use the formula you yourself mentioned to find the length of all 6 line segments, than 4 lengths should be equal (the edges or sides), while to of them must not be equal(the diagonals)

Note that if the length of the diagonals is equal than the shape is a square and not the rhombus