How to find the vertex of a rhombus? I am unable to solve this question.
If the area of a rhombus is 10 sq.unit . It's diagonals intersect at (0,0) if one vertex of the rhombus is (3,4) , then one of the other vertices can be ?
I took a rhombus as ABCD . I took A as (3,4)  and took O(0,0) as the point of the intersections of the diagonals . I found out OA as 5 and OB as 1 . I found out C as(-3,-4)  . Now , the problem is that I am unable to find  the vertex B . Please tell me how do I find out the vertex B . Thank you!
 A: The length of $A C$ is 10, so the length of $B D$ is 2. Now $O C$ and $O B$ are orthogonal, so $B = t (4, -3)$. Putting in the distance you have $t = 1/5$.
A: As we know that the diagonals of rhombus are equal in length & intersect each other normally.  Distance of each vertex from the origin is $\sqrt{3^2+4^2}=5$.   
Thus assuming the rhombus ABCD, the vertex C opposite to $A\equiv (3, 4)$ can be easily determined as the moi-point of AC is $(0, 0)$ Hence, $C(-3, -4)$. Now, assume any vertex say $(x, y)$ on the other diagonal BD. Since, the semi-diagonal OB is normal to OA, we get $$\left(\frac{y-0}{x-0}\right)\left(\frac{4-0}{3-0}\right)=-1 \implies y=-\frac{3}{4}x \tag 1$$ Since $OB=OA=\sqrt{3^2+4^2}=5$, we get 
$$\sqrt{(x-0)^2+(y-0)^2}=5 \implies x^2+y^2=25$$ Setting the value of $y$, we get $$x^2+\left(-\frac{3}{4}x\right)^2=25 \implies x^2=16 \implies x=\pm 4$$$$ x=4\quad  \implies y=-\frac{3}{4}(4)=-3$$ $$ x=-4\quad  \implies y=-\frac{3}{4}(-4)=3$$ Thus, all the unknown vertices of rhombus ABCD are $B\equiv (4, -3)$, $C\equiv(-3, -4)$ & $D\equiv(-4, 3)$ 
