1. ## coordinate geometry-

The points have the following coordinates:
A(3,-2)
B(p,3)
C(6,2)
D(q,r)
The figure ABCD is a rhombus. Find the values of p, q and r.
Even though I used the midpoint formula and the distance formula, I could not deduce the values of p and q.
with the midpoint formula, i did get the value for r (-3).

2. Originally Posted by Ilsa
The points have the following coordinates:
A(3,-2)
B(p,3)
C(6,2)
D(q,r)
The figure ABCD is a rhombus. Find the values of p, q and r.
Even though I used the midpoint formula and the distance formula, I could not deduce the values of p and q.
with the midpoint formula, i did get the value for r (-3).

Hi Ilsa,

First find the slope of AC. That would be $\dfrac{4}{3}$

Since the diagonals of a rhombus are perpendicular, the slope of BD would be $-\dfrac{3}{4}$

Next, find the midpoint of AC. That would be $M\left(\dfrac{9}{2}, 0\right)$

Now, use points $M\left(\dfrac{9}{2}, 0\right)\: \text{and}\: B(p, 3)$ and slope $-\dfrac{3}{4}$ to find the missing x-coordinate p.

$\dfrac{0-3}{\frac{9}{2}-p}=-\dfrac{3}{4}$

This all boils down to $\boxed{p=\dfrac{1}{2}}$

Since, M is also the midpoint of BD, we can easily find the coordinates of D this way:

$\dfrac{q+\frac{1}{2}}{2}=\dfrac{9}{2}$ and $\dfrac{r+3}{2}=0$

Solving the above will yield $\boxed{q=\dfrac{17}{2}}$ and $\boxed{r=-3}$

The coordinates of the vertices of your rhombus are:

$A\3, -2)" alt="A\3, -2)" />

$B\left(\dfrac{1}{2}, 3\right)$

$C\6, 2)" alt="C\6, 2)" />

$D\left(\dfrac{17}{2}, -3\right)$

You can verify that ABCD is a rhombus by using the distance formula:

$AB = BC = CD = AD =\sqrt{31.25}$