Question

A diagonal of rhombus $ABCD$ is member of both the families of lines
$(x+y-1)+{\lambda }_1(2x+3y-2)=0$ and
$(x-y+2)+{\lambda }_2(2x-3y+5)=0$
where ${\lambda }_1$ and ${\lambda }_2\in R$ and one of the vertex of rhombus is $(3,2)$. If area of the rhombus is $12\sqrt{5}$ square units, then the length of the longer diagonal of the rhombus is

Answer 1

Intersection point of $x+y-1=0$ and $2x+3y-2=0$ is $(1,0)$
Intersection point of $x-y+2=0$ and $2x-3y+5=0$ is $(-1,1)$
Since diagonal of the rhombus is a member of both the given families of lines, it will pass through the points $(1,0)$ and $(-1,1)$.
$\Rightarrow$ Equation of one diagonal is $x+2y-1=0.$

Vertex $(3,2)$ does not lie on $x+2y-1=0.$
Finding distance of $(3,2)$ from $x+2y-1=0:$
Distance $=\left|\frac{3+2\times 2-1}{\sqrt{1^2+2^2}}\right|=\frac{6}{\sqrt{5}}$

Let length of one diagonal be $d_1$ and length of the other be $d_2.$
We know that diagonals of a rhombus bisect each other.
$\therefore d_1=2\times \frac{6}{\sqrt{5}}=\frac{12}{\sqrt{5}}$

Given, area $=12\sqrt{5}$
$\Rightarrow \frac{1}{2}\times d_1\times d_2=12\sqrt{5}$
$\Rightarrow d_2=10$