Question

What is the equation of the line that is perpendicular to y=2/3x+4 and that passes through (–2,–2)?

Answer 1

Well, y=2/3x+4  is already in slope-intercept form, therefore    $\bf y=\stackrel{slope}{\cfrac{2}{3}}x+4$.

so, a perpendicular line to a line with that slope, will have a negative reciprocal slope

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{2}{3}\\\\ slope=\cfrac{2}{{{ 3}}}\qquad negative\implies -\cfrac{2}{{{ 3}}}\qquad reciprocal\implies - \cfrac{{{ 3}}}{2}$

so then, what's the equation of a line whose slope is -3/2 and goes through -2,-2?

$\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ -2}}\quad ,&{{ -2}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies -\cfrac{3}{2} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-2)=-\cfrac{3}{2}[x-(-2)] \\\\\\ y+2=-\cfrac{3}{2}(x+2)\implies y+2=-\cfrac{3}{2}x-3\implies y=-\cfrac{3}{2}x-5$