Question

what is an equation of the line that is perpendicular to y + 1 = -3(x-5) and passes through the point (4, 6)

Answer 1

$\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\\text{We have}\ y+1=-3(x-5)\to m_1=-3.\\\\\text{Therefore}\ m_2=-\dfrac{1}{-3}=\dfrac{1}{3}.\\\\\text{The point-slope form:}\ y-y_1=m(x-x_1).\\\\\text{Put the value of slope and the coordinates of the point (4, 6) to the equation}:\\\\\boxed{y-6=\dfrac{1}{3}(x-4)}$