We'll define the point of intersection of the two lines as our origin, which will make this problem a lot easier. The equations of our lines, then, follow: {a₁: y = x√3 {a₂: y = mx, where m is our unknown. Call the origin A. On a₁, the point B will be (1,√3), and the point C on a₂ will be (1,m). From the information given, we determine the following: c = 2 a = |m - √3| b = √(1+m²) We can quickly apply the law of cosines to solve for m: a² = b² + c² - 2bc cos A (m - √3)² = (√(1+m²))² + 2² - 2(2)√(1+m²)(1/7), which reduces to a quadratic equation, 143m² + 98m√3 + 45 = 0 In the process of deriving this equation, we have m√3 + 1 = √(1+m²), so m >= - We'll use the quadratic formula to solve for m: m = (-98√3 ± √(98*98*3 - 4*143*45)/(2*143) = (-98√3 ± 32√3)/286 = or By inspection, we see that < . However, this implies that cos A = -1/7, which means that the supplementary angle of A which is also formed with this intersection, A', exists such that cos A' = 1/7 . Therefore, our two solutions are: and