Question

What is the equation of the line (in slope-intercept form) that passes through the given point and is

perpendicular to the graph of the given equation?
(-5, -6); x + y = 6​

Answer 1

keeping in mind that perpendicular lines have negative reciprocal slopes, hmmmm what's the slope of the equation above anyway?

`\bf x+y=6\implies y = \stackrel{\stackrel{m}{\downarrow }}{-1}x+6\qquad \impliedby \begin{array}c \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill`

`\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{\cfrac{1}{1}\implies 1}}`

so we're really looking for the equation of a line whose slope is 1 and runs through (-5,-6).

`\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{-6})~\hspace{10em} \stackrel{slope}{m}\implies 1 \\\\\\ \begin{array}c \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-6)}=\stackrel{m}{1}[x-\stackrel{x_1}{(-5)}] \\\\\\ y+6=1(x+5)\implies y+6=x+5\implies y=x-1`