(b) \(\frac{\sqrt{17}}{2}\)
Equation of the line through the points (5, 0) and (0, 3)
y – 0 = \(\frac{3-0}{0-5}\) (x - 5)
⇒ y = \(\frac{-3}{5}\)(x - 5)
⇒ 5y + 3x – 15 = 0
∴ Distance of perpendicular from point (4, 4) on the line
5y + 3x – 15 = 0 is \(\bigg|\frac{5\times4+3\times4-15}{\sqrt{5^2+3^2}}\bigg|\)
= \(\frac{|20+12-15|}{\sqrt{25+9}{}}\) = \(\frac{17}{\sqrt{34}}\) units. = \(\frac{\sqrt{17}}{2}\) units.