If the acute angles between the pairs of lines $3 x^{2} - 7 x y + 4 y^{2} = 0$ and $6 x^{2} - 5 x y + y^{2} = 0$ be $θ_{1}$ and $θ_{2}$ respectively, then

$θ_{1}$ = $θ_{2}$

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$θ_{1}$ = 2 $θ_{2}$

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$2 θ_{1}$ = $θ_{2}$

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$θ_{1}$ = 5 $θ_{2}$

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Solution

The correct option is A
$θ_{1}$ = $θ_{2}$

$tan θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}$
$Obviously θ_{1} = t a n^{- 1} (\frac{2 \sqrt{\frac{49}{4} - 12}}{3 + 4}) = t a n^{- 1} (\frac{1}{7}) θ_{2} = t a n^{- 1} (\frac{2 \sqrt{\frac{25}{4} - 6}}{6 + 1}) = t a n^{- 1} (\frac{1}{7})$

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Similar questions

If the acute angles between the pairs of lines $3 x^{2} - 7 x y + 4 y^{2} = 0$ and $6 x^{2} - 5 x y + y^{2} = 0$ be $θ_{1}$ and $θ_{2}$ respectively, then

Q. If the pair of lines

2 x^{2} + 3 x y + y^{2} = 0

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θ_{1}

and

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| tan (θ_{1} - θ_{2}) |

Q. lf the pair of lines

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makes angles

θ_{1}

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tan (θ_{1} - θ_{2}) =

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3 x^{2} - 5 x y + p y^{2} = 0

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p =

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If the acute angles between the pairs of lines 3x2−7xy+4y2=0 and 6x2−5xy+y2=0 be θ1 and θ2 respectively, then

The correct option is A θ1 = θ2 tanθ=2√h2−aba+b Obviously θ1=tan−1(2√494−123+4)=tan−1(17)θ2=tan−1(2√254−66+1)=tan−1(17)

If the acute angles between the pairs of lines $3 x^{2} - 7 x y + 4 y^{2} = 0$ and $6 x^{2} - 5 x y + y^{2} = 0$ be $θ_{1}$ and $θ_{2}$ respectively, then

The correct option is A
$θ_{1}$ = $θ_{2}$

$tan θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}$
$Obviously θ_{1} = t a n^{- 1} (\frac{2 \sqrt{\frac{49}{4} - 12}}{3 + 4}) = t a n^{- 1} (\frac{1}{7}) θ_{2} = t a n^{- 1} (\frac{2 \sqrt{\frac{25}{4} - 6}}{6 + 1}) = t a n^{- 1} (\frac{1}{7})$