The given line is
x – 2y = 3
⇒ y = \(\frac{x}{2}\) − \(\frac{3}{2}\)
∴ Slope, m1 = \(\frac{1}{2}\)
Let ‘m’ be the slope of line AB which passes through (3, 2)
Since the angle between the two lines is 60°.
∴ tan 45° = ± \(\frac{m_2 − m_1}{1 + m_1m_2}\)
⇒ 1 = ± \(\frac{m_2 − \frac{1}{2}}{1 + \frac{1}{2}m_2}\)
⇒ = ± \(\frac{2m_2 − 1}{ m_2 + 2}\)
∴ \(\frac{2m_2 − 1}{ m_2 + 2}\) = 1 or \(\frac{2m_2 − 1}{ m_2 + 2}\) = −1
⇒ m2 = 3 or m2 = − \(\frac{1}{3}\).
∴ Equation of AB is
y – 2 = 3 (x – 3)
⇒ 3x – y = 7 (m2 = 3)
Or,
y − 2 = − \(\frac{1}{3}\)(x − 3) (m2 = − \(\frac{1}{3}\))
⇒ x + 3y = 9