
The equation ${{x}^{2}}+kxy+{{y}^{2}}-5x-7y+6=0$ represents a pair of lines, then $k$ is
A. $\dfrac{5}{3}$
B. $\dfrac{10}{3}$
C. $\dfrac{3}{2}$
D. $\dfrac{3}{10}$
Answer
163.2k+ views
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}}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
${{x}^{2}}+kxy+{{y}^{2}}-5x-7y+6=0\text{ }...(1)$
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
$a=1;h=\dfrac{k}{2};b=1;g=\dfrac{-5}{2};f=\dfrac{-7}{2};c=6$
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
$\begin{align}
& abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
& \Rightarrow (1)(1)(6)+2(\dfrac{-7}{2})(\dfrac{-5}{2})(\dfrac{k}{2})-(1){{\left( \dfrac{-7}{2} \right)}^{2}}-(1){{\left( \dfrac{-5}{2} \right)}^{2}}-(6){{\left( \dfrac{k}{2} \right)}^{2}}=0 \\
& \Rightarrow 6+\dfrac{35k}{4}-\dfrac{49}{4}-\dfrac{25}{4}-\dfrac{3{{k}^{2}}}{2}=0 \\
& \Rightarrow 6{{k}^{2}}-35k+50=0 \\
& \Rightarrow k=\dfrac{10}{3};k=\dfrac{5}{2} \\
\end{align}$
Thus, the value is $k=\dfrac{10}{3}$.
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.
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 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
${{x}^{2}}+kxy+{{y}^{2}}-5x-7y+6=0\text{ }...(1)$
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
$a=1;h=\dfrac{k}{2};b=1;g=\dfrac{-5}{2};f=\dfrac{-7}{2};c=6$
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
$\begin{align}
& abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
& \Rightarrow (1)(1)(6)+2(\dfrac{-7}{2})(\dfrac{-5}{2})(\dfrac{k}{2})-(1){{\left( \dfrac{-7}{2} \right)}^{2}}-(1){{\left( \dfrac{-5}{2} \right)}^{2}}-(6){{\left( \dfrac{k}{2} \right)}^{2}}=0 \\
& \Rightarrow 6+\dfrac{35k}{4}-\dfrac{49}{4}-\dfrac{25}{4}-\dfrac{3{{k}^{2}}}{2}=0 \\
& \Rightarrow 6{{k}^{2}}-35k+50=0 \\
& \Rightarrow k=\dfrac{10}{3};k=\dfrac{5}{2} \\
\end{align}$
Thus, the value is $k=\dfrac{10}{3}$.
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.
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