
If the equation $\lambda {{x}^{2}}+2{{y}^{2}}-5xy+5x-7y+3=0$ represents two straight lines, then the value of $\lambda $ will be
A. $3$
B. $2$
C. $8$
D. $-8$
Answer
164.1k+ views
Hint: In this question, we are to find the value of $\lambda $. Since the given equation represents two straight lines, we can apply the pair of lines condition and on simplifying, we get the required value of the given equation.
Formula Used:The equation of the pair of straight lines is
$H\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}=0$
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 $S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ 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
$\lambda {{x}^{2}}+2{{y}^{2}}-5xy+5x-7y+3=0\text{ }...(1)$
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
$a=\lambda ;h=\dfrac{-5}{2};b=2;g=\dfrac{5}{2};f=\dfrac{-7}{2};c=3$
If the given equation (1) represents two pairs of lines, then
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0\text{ }...(3)$
On substituting the above values in (3), we get
$\begin{align}
& abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
& \Rightarrow (\lambda )(2)(3)+2(\dfrac{-7}{2})(\dfrac{5}{2})(\dfrac{-5}{2})-{{\left( \dfrac{-7}{2} \right)}^{2}}(\lambda )-{{\left( \dfrac{5}{2} \right)}^{2}}(2)-(3){{\left( \dfrac{-5}{2} \right)}^{2}}=0 \\
& \Rightarrow 6\lambda +\dfrac{175}{4}-\dfrac{49\lambda }{4}-\dfrac{50}{4}-\dfrac{75}{4}=0 \\
& \Rightarrow \dfrac{24\lambda -49\lambda }{4}+\dfrac{25}{2}=0 \\
& \Rightarrow \dfrac{25\lambda }{4}=\dfrac{25}{2} \\
& \therefore \lambda =2 \\
\end{align}$
Thus, the value is $\lambda =2$.
Option ‘B’ is correct
Note: Here, the given equation represents pair of lines. So, the given equation should satisfy the condition we have i.e., $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$. 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.
Formula Used:The equation of the pair of straight lines is
$H\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}=0$
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 $S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ 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
$\lambda {{x}^{2}}+2{{y}^{2}}-5xy+5x-7y+3=0\text{ }...(1)$
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
$a=\lambda ;h=\dfrac{-5}{2};b=2;g=\dfrac{5}{2};f=\dfrac{-7}{2};c=3$
If the given equation (1) represents two pairs of lines, then
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0\text{ }...(3)$
On substituting the above values in (3), we get
$\begin{align}
& abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
& \Rightarrow (\lambda )(2)(3)+2(\dfrac{-7}{2})(\dfrac{5}{2})(\dfrac{-5}{2})-{{\left( \dfrac{-7}{2} \right)}^{2}}(\lambda )-{{\left( \dfrac{5}{2} \right)}^{2}}(2)-(3){{\left( \dfrac{-5}{2} \right)}^{2}}=0 \\
& \Rightarrow 6\lambda +\dfrac{175}{4}-\dfrac{49\lambda }{4}-\dfrac{50}{4}-\dfrac{75}{4}=0 \\
& \Rightarrow \dfrac{24\lambda -49\lambda }{4}+\dfrac{25}{2}=0 \\
& \Rightarrow \dfrac{25\lambda }{4}=\dfrac{25}{2} \\
& \therefore \lambda =2 \\
\end{align}$
Thus, the value is $\lambda =2$.
Option ‘B’ is correct
Note: Here, the given equation represents pair of lines. So, the given equation should satisfy the condition we have i.e., $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$. 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.
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