The angle between the pair of straight lines $x^{2} + 4 y^{2} - 7 x y = 0,$ is

$t a n^{- 1} \frac{1}{3}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$t a n^{- 1} 3$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$t a n^{- 1} \frac{\sqrt{33}}{5}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$t a n^{- 1} \frac{5}{\sqrt{33}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$t a n^{- 1} \frac{\sqrt{33}}{5}$

$tan θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}$
$θ = t a n^{- 1} \frac{2 \sqrt{\frac{49}{4} - 4}}{5} = t a n^{- 1} \frac{\sqrt{33}}{5}$ .

Suggest Corrections

Similar questions

Q. Find the equations of the straight lines bisecting the angles between the pairs of straight lines :

12 x^{2} - 7 x y - 12 y^{2} = 0; 71 x^{2} + 94 x y - 71 y^{2} = 0; x^{2} - y^{2} = 0

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. The sine of the angle between the pair of lines represented by the equation

x^{2} - 7 x y + 12 y^{2} = 0

Q. Find the acute angle between the pair of lines represented by the following equation.

x^{2} - 7 x y + 12 y^{2} = 0

Q. The angle between the pair of straight lines

x^{2} - y^{2} - 2 y - 1 = 0

Join BYJU'S Learning Program

The angle between the pair of straight lines x2+4y2−7xy=0, is

The correct option is C tan−1√335 tanθ=2√h2−aba+b θ=tan−12√494−45=tan−1√335.

The angle between the pair of straight lines $x^{2} + 4 y^{2} - 7 x y = 0,$ is

The correct option is C
$t a n^{- 1} \frac{\sqrt{33}}{5}$

$tan θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}$
$θ = t a n^{- 1} \frac{2 \sqrt{\frac{49}{4} - 4}}{5} = t a n^{- 1} \frac{\sqrt{33}}{5}$ .