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