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 = \(\frac{18}{5}\)
\(\frac{10 + 5}{5}\) = c
y = -\(\frac{2}{5}\)x + \(\frac{18}{5}\)
5y = -2x + 18
or 5y + 2x - 18 = 0