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Question: Derive the formula for the tangent of the acute angle between a pair of straight lines represented b...

Derive the formula for the tangent of the acute angle between a pair of straight lines represented by the general homogeneous equation of the second degree ax2+2hxy+by2=0ax^2 + 2hxy + by^2 = 0.

Answer

The formula for the tangent of the acute angle ϕ\phi between the pair of straight lines ax2+2hxy+by2=0ax^2 + 2hxy + by^2 = 0 is: tan⁡ϕ=2h2−ab∣a+b∣\tan \phi = \frac{2\sqrt{h^2 - ab}}{|a+b|} This formula is valid for real lines, i.e., when h2−ab≥0h^2 - ab \ge 0.

Explanation

Solution

The general equation of a pair of straight lines passing through the origin is given by: ax2+2hxy+by2=0ax^2 + 2hxy + by^2 = 0 If b≠0b \neq 0, we can divide by x2x^2 to get a quadratic equation in terms of the slope m=y/xm = y/x: b(yx)2+2h(yx)+a=0b\left(\frac{y}{x}\right)^2 + 2h\left(\frac{y}{x}\right) + a = 0 Let m1m_1 and m2m_2 be the slopes of the two lines. From Vieta's formulas applied to the quadratic equation bm2+2hm+a=0bm^2 + 2hm + a = 0:

  • Sum of slopes: m1+m2=−2hbm_1 + m_2 = -\frac{2h}{b}
  • Product of slopes: m1m2=abm_1 m_2 = \frac{a}{b}

Let θ\theta be the angle between the two lines. The tangent of the angle is given by: tan⁡θ=∣m1−m21+m1m2∣\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| To find ∣m1−m2∣|m_1 - m_2|, we first find (m1−m2)2(m_1 - m_2)^2: (m1−m2)2=(m1+m2)2−4m1m2(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2 Substituting the values from Vieta's formulas: (m1−m2)2=(−2hb)2−4(ab)=4h2b2−4ab=4h2−4abb2=4(h2−ab)b2(m_1 - m_2)^2 = \left(-\frac{2h}{b}\right)^2 - 4\left(\frac{a}{b}\right) = \frac{4h^2}{b^2} - \frac{4a}{b} = \frac{4h^2 - 4ab}{b^2} = \frac{4(h^2 - ab)}{b^2} Taking the square root, we get: ∣m1−m2∣=4(h2−ab)b2=2h2−ab∣b∣|m_1 - m_2| = \sqrt{\frac{4(h^2 - ab)}{b^2}} = \frac{2\sqrt{h^2 - ab}}{|b|} For real lines, h2−ab≥0h^2 - ab \ge 0. Now, substitute ∣m1−m2∣|m_1 - m_2| and 1+m1m21 + m_1 m_2 into the tangent formula: tan⁡θ=∣2h2−ab∣b∣1+ab∣=∣2h2−ab∣b∣b+ab∣\tan \theta = \left| \frac{\frac{2\sqrt{h^2 - ab}}{|b|}}{1 + \frac{a}{b}} \right| = \left| \frac{\frac{2\sqrt{h^2 - ab}}{|b|}}{\frac{b+a}{b}} \right| tan⁡θ=2h2−ab∣b∣⋅∣b∣∣a+b∣=2h2−ab∣a+b∣\tan \theta = \frac{2\sqrt{h^2 - ab}}{|b|} \cdot \frac{|b|}{|a+b|} = \frac{2\sqrt{h^2 - ab}}{|a+b|} This formula gives the tangent of the acute angle between the pair of straight lines. The absolute value in the denominator ensures that the value is positive, corresponding to the acute angle.