Find the equation of the line that contains the point (βˆ’3, 8) and that is perpendicular to a line that passes through the points (βˆ’6, 2) and (6, 4)

Accepted Solution

To arrive at the desired answer, we need a point on the line and the slope of the line. A point on the line is already given: (-3,8). As for the slope, we first compute the slope of the line passing through (-6,2) and (6,4): $$m_1 = \dfrac{2-4}{-6-6} = \dfrac{-2}{-12} = \dfrac{1}{6}.$$ The required line is perpendicular to the given line, so the slope of the required line should be the negative reciprocal of the slope of the given line: $$m = -\dfrac{1}{m_1} = -\dfrac{1}{\dfrac{1}{6}} = -6.$$ So the required line has slope -6 and contains (-3,8). Thus, the point-slope form of the equation of the line is $$y-8 = -6(x-(-3)) \Longrightarrow \boxed{y-8 = -6(x+3).}$$