# Getting a rhombus vertex from one vertex and its area

1. Jun 9, 2010

### Telemachus

Hi there. Im tryin to solve this one, I know that if I find the way to get one more vertex i'd have it solved.

1. The problem statement, all variables and given/known data
The point A(-1,8) is the vertex of a rhombus which minor diagonal is situated on the line L: $$L=\begin{Bmatrix} x=3\mu & \mbox{ }& \\y=1+4\mu & \mbox{}&\end{matrix}$$, $$\mu\in{R}$$. Get the coordinates of the rest of the vertex knowing that the rhombus area is 30.

2. Relevant equations
$$A=\displaystyle\frac{dD}{2}$$

3. The attempt at a solution
Well, I haven't done too much. Actually I did some, but then I realized that I had confused something, cause I got the line L' where would be located the major D, but I've used for it the point A, and then I was trying to get the point of intersection between L and L', so then I doubled it, and I was going to get my second vertex, but then I realized that it was wrong, cause I couln't use point A, cause A don't belongs to L', it belongs to L, so...

$$30=\displaystyle\frac{dD}{2}\Rightarrow{60=dD}$$

So, I don't know much about L', but that its perpendicular to L.

$$L'=\begin{Bmatrix} x=x_0+4/3\lambda & \mbox{ }& \\y=y_0-\lambda & \mbox{}&\end{matrix}$$

Last edited: Jun 9, 2010
2. Jun 9, 2010

### tiny-tim

Hi Telemachus!
That's right , but from then on you're making it too complicated.

Hint: Where is the midpoint of the rhombus?

3. Jun 9, 2010

### Telemachus

On the intersection between L and L', but how should I get L'? I thought maybe using pythagoric equation I could find another vertex, but im not too sure. I know it gives enough data, but I don't know how to use it.

4. Jun 9, 2010

### Telemachus

I had a misunderstood, I think I was on the right way, cause actually I didn't corroborate if A belongs to L, and the sentence don't say so. I'll corroborate, if it doesn't I was on the right way, and it will be easy to solve, cause if A don't belongs to L, it must belongs to L'.

Bye there, and thanks.

5. Jun 9, 2010

### tiny-tim

L' is perpendicular to L, and goes through A.

(and remember, you don't need to find L', you only need the midpoint )

6. Jun 9, 2010

### Telemachus

How could I get the midpoint without L'? I thought of it as the intersection between L and L'.

7. Jun 9, 2010

### tiny-tim

AM is on L', so it's perpendicular to L.