2y = 5x + 4 (4, 2) y = \(\frac{5x}{2}\) + 4 comparing with y = mx + e m = \(\frac{5}{2}\) Since they are perpendicular m1m2 = -1 m2 = \(\frac{-1}{m_1}\) = -1 \(\frac{5}{2}\) = -1 x \(\frac{2}{5}\) The equator of the line is thus y = mn + c (4, 2) 2 = -\(\frac{2}{5}\)(4) + c \(\frac{2}{1}\) + \(\frac{8}{5}\) = c c = \(\frac{18}{5}\) y = -\(\frac{2}{5}\)x + \(\frac{18}{5}\) 5y = -2x + 18 or 5y + 2x - 18 = 0