Question

Area of Rhombus is 10 sq unit . It's diagonals intersect at $(0,0)$, if one vertex of rhombus is $(3,4)$, then one of the other vertices is:

• A. (4,-3)
• B. (4,2)
• C. (3,4)
• D. None of these

Answer 1

Answer: (A)

(4,-3)

Given,

Area of rhombus $=10$

one vertex of rhombus $=(3,4)$

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

$\left(\frac{y-0}{x-0}\right)\left(\frac{4-0}{3-0}\right)=-1$

$\Rightarrow y=-\frac{3}{4}x$

As we know that the diagonals of rhombus are equal in length & intersect each other normally. Distance of each vertex from the origin is $\sqrt{3^2+4^2}=5$

Since $OA=OB=\sqrt{3^2+4^2}=5$

$\sqrt{(x-0{)}^2+(y-0{)}^2}=5$

$\Rightarrow x^2+y^2=25$

substituting the value of $y$, we get,

$x^2+{(-\frac{3}{4}x)}^2=25$

$x^2=16$

$x=\pm 4$

$\to y=\mp 3$

Thus, all the unknown vertices of rhombus ABCD are $B\equiv (4,-3),C\equiv (-3,-4),D\equiv (-4,3)$