# Getting a rhombus vertex knowing one vertex and its area

• June 9th 2010, 07:22 AM
Ulysses
Getting a rhombus vertex knowing one vertex and its area
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.

The point A(-1,8) is the vertex of a rhombus which minor diagonal is situated on the line L: $L=\left\{\begin{array}{cc}x=3\mu,&\mbox{ } \\y=1+4\mu, & \mbox{ }\end{array}\right.$

, $\mu\in{R}$. Get the coordinates of the rest of the vertex knowing that the rhombus area is 30.

$A=\displaystyle\frac{dD}{2}$

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.

$L'=\left\{\begin{array}{cc}x=x_0+4/3\lambda,&\mbox{ } \\y=y_0-\lambda, & \mbox{ }\end{array}\right.$
• June 9th 2010, 07:34 AM
earboth
Quote:

Originally Posted by Ulysses
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.

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 \\y=1+4\mu \end{Bmatrix}
$

$L=\left\{\begin{array}{cc}x=3\mu\\y=1+4\mu \end{array}\right\}$, $\mu\in{R}$. Get the coordinates of the rest of the vertex knowing that the rhombus area is 30.

$A=\displaystyle\frac{dD}{2}$

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{Bmatrix}
$