Question

In a rhombus $ABCD,$ the slope of diagonal $AC$ 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 $(-2,3)$ and the area of rhombus is $10\sqrt{2}$ sq. units, then which of the following is/are correct?

• A. Length of smaller diagonal is $5$
• B. Length of larger diagonal is $6\sqrt{2}$
• C. One of the vertex is $(2,-1)$
• D. Perimeter of rhombus is $2\sqrt{57}$ units

Answer 1

Answer: (A), (C), (D)

The correct options are
A Length of smaller diagonal is $5$
C One of the vertex is $(2,-1)$
D Perimeter of rhombus is $2\sqrt{57}$ units

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 $AC=1,$ therefore equation of $AC:x-y+1=0$
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 $(0,1)$
$\Rightarrow B:(x,y)=(2,-1)$
$BD=4\sqrt{2}=d_2$

Area of rhombus $=10\sqrt{2}$
$\Rightarrow \frac{1}{2}(d_1\times d_2)=10\sqrt{2}$
$\Rightarrow d_1=5$

Side of rhombus $=\sqrt{\frac{d_1^2}{4}+\frac{d_2^2}{4}}=\frac{\sqrt{57}}{2}$
$\therefore$ Perimeter $=2\sqrt{57}$