Question

Find the equation of the line that is perpendicular to the given line and passes through the given point. Enter the right side of the equation as as a single fraction. y=3x-12/2; (9,9)

Answer 1

$Let\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\\\\\text{We have}\ k:y=\dfrac{3x-12}{2}=\dfrac{3x}{2}-\dfrac{12}{2}=\dfrac{3}{2}x-6\to m_1=\dfrac{3}{2}\\\\l:y=mx+b\to\dfrac{3}{2}m=-1\qquad|\cdot\dfrac{2}{3}\\\\m=-\dfrac{2}{3}\\\\l:y=-\dfrac{2}{3}x+b\\\\\text{Put the coordinates of the point (9, 9) to the equation of the line}\ l:\\\\9=-\dfrac{2}{3}(9)+b\\\\9=-2(9)+b\\\\9=-18+b\qquad|+18\\\\b=27\\\\y=-\dfrac{2}{3}x+27\to y=-\dfrac{2x}{3}+\dfrac{27\cdot3}{3}\to\boxed{y=\dfrac{-2x+81}{3}}$