Question

The value of $k$, so that the equation $2{{x}^{2}}+5xy+3{{y}^{2}}+6x+7y+k=0$ represents a pair of straight lines, is
A. $4$
B. $6$
C. $0$
D. $8$

Answer 1

Hint: In this question, we need to find the value of $k$ 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 ${{h}^{2}}=ab$, 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
$2{{x}^{2}}+5xy+3{{y}^{2}}+6x+7y+k=0\text{ }...(1)$
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
$a=2;h=\dfrac{5}{2};b=3;g=3;f=\dfrac{7}{2};c=k$
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
$\begin{align}
  & abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
 & \Rightarrow (2)(3)(k)+2(\dfrac{7}{2})(3)(\dfrac{5}{2})-2{{\left( \dfrac{7}{2} \right)}^{2}}-(3){{\left( 3 \right)}^{2}}-(k){{\left( \dfrac{5}{2} \right)}^{2}}=0 \\
 & \Rightarrow 6k+\dfrac{105}{2}-\dfrac{49}{2}-27-\dfrac{25k}{4}=0 \\
 & \Rightarrow 6k-\dfrac{25k}{4}=-\dfrac{105}{2}+\dfrac{49}{2}+27 \\
 & \Rightarrow \dfrac{-1k}{4}=-1 \\
 & \therefore k=4 \\
\end{align}$
Thus, the value is $k=4$.



Option ‘A’ 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.