ANGLE BETWEEN THE PAIR OF LINES

Let y = m1 x and y = m2 x be the lines represented by ax2 + 2hxy + by2 = 0. Then, $${{m}_{1}}+{{m}_{2}}=\frac{-2h}{b}\,$$ and $${{m}_{1}}{{m}_{2}}=\frac{a}{b}$$ … (i)Let θ be the angle between the lines y = m1 x and y = m2 x. Then,

⇒ $$\tan \theta =\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}$$.

⇒ $$\tan \theta =\frac{\sqrt{{{\left( {{m}_{1}}+{{m}_{2}} \right)}^{2}}-4{{m}_{1}}{{m}_{2}}}}{1+{{m}_{1}}{{m}_{2}}}$$.

⇒ $$\tan \theta =\frac{\sqrt{\frac{4{{h}^{2}}}{{{b}^{2}}}-\frac{4a}{b}}}{1+\frac{a}{b}}$$  [Using (i)]

⇒ $$\tan \theta =\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}$$,

⇒ $$\theta ={{\tan }^{-1}}\left\{ \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right\}$$.

Thus, the angle θ between the pair of lines represented by ax2 + 2hxy + by2 = 0 is given by $$\tan\theta =\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}$$.

PROBLEM: Prove that the angle between the straight lines given by (x cosα – y sinα)2 = (x2 + y2) sin²α is 2α.

SOLUTION: We have,

(x cos α – y sin α)2 = (x2 + y2) sin2 α

⇒ x2 (cos2 α – sin2 α) – xy sin 2α + 0.y² = 0.

Comparing this equation with ax² + 2hxy + by² = 0, we get

a = cos²α – sin² α, b = 0 and 2h = – sin 2α

Let θ be the angle between the given lines. Then,

⇒ $$\tan \theta =\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}$$,

⇒ $$\tan \theta =\frac{2\sqrt{\frac{\left( {{\sin }^{2}}2\alpha \right)}{4}-0}}{{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha }$$,

⇒ $$\tan \theta =\frac{\sin 2\alpha }{\cos 2\alpha }$$,

⇒ θ = 2α