- #1
Telemachus
- 835
- 30
Hi there. I am 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: [tex]L=\begin{Bmatrix} x=3\mu & \mbox{ }& \\y=1+4\mu & \mbox{}&\end{matrix}
[/tex], [tex]\mu\in{R}[/tex]. Get the coordinates of the rest of the vertex knowing that the rhombus area is 30.
[tex]A=\displaystyle\frac{dD}{2}[/tex]
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...
[tex]30=\displaystyle\frac{dD}{2}\Rightarrow{60=dD}[/tex]
So, I don't know much about L', but that its perpendicular to L.
[tex]L'=\begin{Bmatrix} x=x_0+4/3\lambda & \mbox{ }& \\y=y_0-\lambda & \mbox{}&\end{matrix}
[/tex]
Homework Statement
The point A(-1,8) is the vertex of a rhombus which minor diagonal is situated on the line L: [tex]L=\begin{Bmatrix} x=3\mu & \mbox{ }& \\y=1+4\mu & \mbox{}&\end{matrix}
[/tex], [tex]\mu\in{R}[/tex]. Get the coordinates of the rest of the vertex knowing that the rhombus area is 30.
Homework Equations
[tex]A=\displaystyle\frac{dD}{2}[/tex]
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...
[tex]30=\displaystyle\frac{dD}{2}\Rightarrow{60=dD}[/tex]
So, I don't know much about L', but that its perpendicular to L.
[tex]L'=\begin{Bmatrix} x=x_0+4/3\lambda & \mbox{ }& \\y=y_0-\lambda & \mbox{}&\end{matrix}
[/tex]
Last edited: