Question

$ABCD$ is a rhombus. The slope of $AC$ is $1$ and is among the family of lines $(x+2y-5)+\lambda (3x+y-5)=0,$ where $\lambda \in R.$. One of the vertex of the rhombus is $(-2,3)$. If the area of rhombus is $10\sqrt{2}$ sq. units, then which of the following is/are correct?

• A. Length of smaller diagonal is $4$
• B. Length of larger diagonal is $4\sqrt{2}$
• C. One of the vertex is $(2,-1)$
• D. Perimeter of rhombus is $2\sqrt{57}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B Length of larger diagonal is $4\sqrt{2}$
C One of the vertex is $(2,-1)$
D Perimeter of rhombus is $2\sqrt{57}$

The diagonal $AC$ is the member of family of line $(x+2y-5)+\lambda (3x+y-5)=0,$ so it will pass through the intersection point of lines $x+2y-5=0$ and $3x+y-5=0$ i.e., $(1,2).$
Since, slope of $AC$ is $1$, therefore equation of
$AC$ is, $x-y+1=0$
The point $(-2,3)$ does not lie on $x-y+1=0$, so it will lie on another diagonal $BD$.
Since, diagonals of rhombus are perpendicular to each other, therefore equation of $BD$ is, $x+y-1=0$
Point of intersection of $AC$ and $BD$ is $(0,1)$.

$\Rightarrow B:(x,y)=(2,-1)$
$BD=4\sqrt{2}=d_2$
$\text{Area of rhombus}=\frac{1}{2}(d_1\times d_2)=10\sqrt{2}$
$\Rightarrow d_1=5$
$\text{Side of rhombus}=\sqrt{\frac{d_1^2}{4}+\frac{d_2^2}{4}}=\frac{\sqrt{57}}{2}$
$\text{Perimeter}=2\sqrt{57}$