
If the acute angle between the pairs of lines $3{x^2} - 7xy + 4{y^2} = 0$ and $6{x^2} - 5xy + {y^2} = 0$ be ${\theta _1}$ and ${\theta _2}$ respectively, then
A ${\theta _1} = {\theta _2}$
B ${\theta _1} = 2{\theta _2}$
C $2{\theta _1} = {\theta _2}$
D None of these
Answer
163.5k+ views
Hint: First we will compare both given equations with the general equation to find the value of $a,\,b,\,h$. After finding the values will put in the formula to find the angle between the lines. After finding the angle will find the relation between both the angles.
Formula Used: $\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Complete step by step solution: $3{x^2} - 7xy + 4{y^2} = 0$
The general equation is $a{x^2} + 2hxy + b{y^2} = 0$
On comparing, we will get
$a = 3$, $h = \dfrac{{ - 7}}{2}$ and $b = 4$
We know the angle between the two lines is
$\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
$\tan {\theta _1} = \left| {\dfrac{{2\sqrt {{{( - \dfrac{7}{2})}^2} - (3)(4)} }}{{3 + 4}}} \right|$
After solving, we will get
$\tan {\theta _1} = \left| {\dfrac{{2\sqrt {\dfrac{{49}}{4} - 12} }}{7}} \right|$
$\tan {\theta _1} = \left| {\dfrac{{2\sqrt {\dfrac{1}{4}} }}{7}} \right|$
After solving, we get
$\tan {\theta _1} = \dfrac{1}{7}$
${\theta _1} = {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right)$ …(1)
$6{x^2} - 5xy + {y^2} = 0$
On comparing with general equation, we will get
$a = 6$, $h = \dfrac{{ - 5}}{2}$ and $b = 1$
We know the angle between the two lines is
$\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
$\tan {\theta _2} = \left| {\dfrac{{2\sqrt {{{( - \dfrac{5}{2})}^2} - (6)(1)} }}{{6 + 1}}} \right|$
After solving, we get
$\tan {\theta _2} = \left| {\dfrac{{2\sqrt {\dfrac{{25}}{4} - 6} }}{7}} \right|$
$\tan {\theta _2} = \left| {\dfrac{{2\sqrt {\dfrac{1}{4}} }}{7}} \right|$
After solving, we get
$\tan {\theta _2} = \dfrac{1}{7}$
${\theta _2} = {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right)$ …(2)
From equation (1) and equation (2)
${\theta _1} = {\theta _2}$
Hence, option A is correct.
Note: Students should use correct formulas of angle between the two lines to get the correct answer. While comparing with the general equation they should carefully find the value of $a,\,b,\,h$ to avoid any further calculation errors.
Formula Used: $\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Complete step by step solution: $3{x^2} - 7xy + 4{y^2} = 0$
The general equation is $a{x^2} + 2hxy + b{y^2} = 0$
On comparing, we will get
$a = 3$, $h = \dfrac{{ - 7}}{2}$ and $b = 4$
We know the angle between the two lines is
$\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
$\tan {\theta _1} = \left| {\dfrac{{2\sqrt {{{( - \dfrac{7}{2})}^2} - (3)(4)} }}{{3 + 4}}} \right|$
After solving, we will get
$\tan {\theta _1} = \left| {\dfrac{{2\sqrt {\dfrac{{49}}{4} - 12} }}{7}} \right|$
$\tan {\theta _1} = \left| {\dfrac{{2\sqrt {\dfrac{1}{4}} }}{7}} \right|$
After solving, we get
$\tan {\theta _1} = \dfrac{1}{7}$
${\theta _1} = {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right)$ …(1)
$6{x^2} - 5xy + {y^2} = 0$
On comparing with general equation, we will get
$a = 6$, $h = \dfrac{{ - 5}}{2}$ and $b = 1$
We know the angle between the two lines is
$\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
$\tan {\theta _2} = \left| {\dfrac{{2\sqrt {{{( - \dfrac{5}{2})}^2} - (6)(1)} }}{{6 + 1}}} \right|$
After solving, we get
$\tan {\theta _2} = \left| {\dfrac{{2\sqrt {\dfrac{{25}}{4} - 6} }}{7}} \right|$
$\tan {\theta _2} = \left| {\dfrac{{2\sqrt {\dfrac{1}{4}} }}{7}} \right|$
After solving, we get
$\tan {\theta _2} = \dfrac{1}{7}$
${\theta _2} = {\tan ^{ - 1}}\left( {\dfrac{1}{7}} \right)$ …(2)
From equation (1) and equation (2)
${\theta _1} = {\theta _2}$
Hence, option A is correct.
Note: Students should use correct formulas of angle between the two lines to get the correct answer. While comparing with the general equation they should carefully find the value of $a,\,b,\,h$ to avoid any further calculation errors.
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