If the angle between two lines represented by $2 x^{2} + 5 x y + 3 y^{2} + 7 y + 4 = 0$ is ${tan}^{- 1} m$ , then $m$ is equal to.

\frac{1}{5}

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1

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\frac{7}{5}

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7

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Solution

The correct option is A $\frac{1}{5}$
${2 x}^{2} + 5 x y + {3 y}^{2} + 7 y + 4 = 0$
Comparing above eqn. with general eqn. of pair of straight line
${a x}^{2} + 2 h x y + {b y}^{2} + 2 g x + 2 f y + c = 0$
We get $a = 2,$ $h = \frac{5}{2},$ $b = 3,$ $g = 0,$ $f = \frac{7}{2}$ and $c = 4$
Angle b/w pair of straight lines is
$θ = {tan}^{- 1} [\frac{2 \sqrt{h^{2} - a b}}{a + b}]$
${tan}^{- 1} m = {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{2 \sqrt{\frac{25}{4} - 6}}{2 + 3} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$m = \frac{\sqrt{25 - 24}}{5}$
$m = \frac{1}{5}$

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Similar questions

If the angle between the two lines represented by $2 x^{2} + 5 x y + 3 y^{2} + 6 x + 7 y + 4 = 0$ is ${tan}^{- 1} m$ , then m =

2 x^{2} + 5 x y + 3 y^{2} + 6 x + 7 y + 4 = 0

{tan}^{- 1}

2 x^{2} + 5 x y + 3 y^{2} + 6 x + 7 y + 4 = 0

{tan}^{- 1} (m),

m

2 x^{2} + 5 x y + 3 y^{2} + 6 x + 7 y + 4 = 0

{tan}^{- 1} (m)

m

2 x^{2} + 5 x y + 3 y^{2} + 6 x + 7 y + 4 = 0

{tan}^{- 1} (m)

m = \frac{1}{5}

θ

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

{tan}^{- 1} \frac{(2 \sqrt{h^{2} - a b})}{| a + b |}

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