Question

Find the acute angle (in degree) between the pair of straight lines $4x^2+14xy+6y^2=0$

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Answer 1

Given equation is $4x^2+14xy+6y^2=0$ Comparing this equation with

$a{x}^2+2hxy+b{y}^2=0$

a=4, h=7, b=6

Now,

We know, $\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

Substituting the value of a, b and h

= $\left|\frac{2\sqrt{49-4\times 6}}{4+6}\right|$

= $\frac{2\times 5}{10}$

$tan\theta =1or-1$

$\theta ={45}^{\circ }$ and ${135}^{\circ }$

Acute angle between the lines is ${45}^{\circ }$ and obtuse angle between them is ${135}^{\circ }$