ok two steps to this
first we find the slope of the line passing through the two points
\(m = \dfrac{4-1}{4-6} = -\dfrac{3}{2}\)
now we want a line perpendicular to this so we have a slope of
\(m_{\perp} = -\dfrac{1}{m} = -\left(\dfrac{1}{-\frac 3 2}\right) = \dfrac 2 3\)
now using point slope formula with the point (-2,6) we have
\((y-6) = \dfrac 2 3 (x - (-2)) \\ y = \dfrac 2 3 x + \dfrac{22}{3} \)
.
