The value of $\lambda $ for which the equation ${{x}^{2}}-\lambda xy+2{{y}^{2}}+3x-5y+2=0$ may represents a pair of straight lines is
A. $2$
B. $3$
C. $4$
D. $1$

Hint: In this question, we need to find the value of $\lambda $ in the equation that represents two straight lines. So, we can apply the formula $\Delta =0$ to find the required value.



Formula Used:The equation of the pair of straight lines is written as
$H\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}=0$
This is called a homogenous equation of the second degree in $x$ and $y$
And
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
This is called a general equation of the second degree in $x$ and $y$.
If ${{h}^{2}}If ${{h}^{2}}=ab$, then $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents coincident lines.
If ${{h}^{2}}>ab$, then $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two real and different lines that pass through the origin.
Thus, the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two lines. They are:
$ax+hy\pm y\sqrt{{{h}^{2}}-ab}=0$
If $S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ represents a pair of lines, then
i) $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$ and
ii) ${{h}^{2}}\ge ab,{{g}^{2}}\ge ac,{{f}^{2}}\ge bc$



Complete step by step solution:Given equation is
${{x}^{2}}-\lambda xy+2{{y}^{2}}+3x-5y+2=0\text{ }...(1)$
But we have the general equation of pair lines as
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\text{ }...(2)$
Comparing (1) and (2), we get
$a=1;h=\dfrac{-\lambda }{2};b=2;g=\dfrac{3}{2};f=\dfrac{-5}{2};c=2$
If the given equation (1) represents two pairs of lines, then
$\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0\text{ }...(3)$
On substituting the above values in (3), we get
$\begin{align}
  & abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
 & \Rightarrow (1)(2)(2)+2(\dfrac{-5}{2})(\dfrac{3}{2})(\dfrac{-\lambda }{2})-(1){{\left( \dfrac{-5}{2} \right)}^{2}}-2{{\left( \dfrac{3}{2} \right)}^{2}}-(2){{\left( \dfrac{-\lambda }{2} \right)}^{2}}=0 \\
 & \Rightarrow 4+\dfrac{15\lambda }{4}-\dfrac{25}{4}-\dfrac{9}{2}-\dfrac{{{\lambda }^{2}}}{2}=0 \\
 & \Rightarrow \dfrac{15\lambda }{4}-\dfrac{{{\lambda }^{2}}}{2}-\dfrac{27}{4}=0 \\
 & \Rightarrow 15\lambda -2{{\lambda }^{2}}-27=0 \\
 & \therefore \lambda =3,\lambda =\dfrac{9}{2} \\
\end{align}$
Thus, the value is $\lambda =3$.



Option ‘B’ is correct

Note: Here, the given equation represents pair of lines. So, the given equation should satisfy the condition we have $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$. Then, by substituting the values into this condition, we get the required values. In this problem, we need to find the coefficient of ${{x}^{2}}$ in the given equation. So, the we applied above formula. On simplifying, we get the required value.