Question

Write the equation described in each case.

The line perpendicular to y =
a. slope of the new line:
b. equation:
x + 6 through (3, 2).

Answer 1

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

`y=\stackrel{\stackrel{m}{\downarrow }}{1}x+6\qquad \impliedby \begin{array}ll \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}`

so a perpendicular line to that one will have

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{1\implies \cfrac{1}{\underline{1}}} ~\hfill \stackrel{reciprocal}{\cfrac{\underline{1}}{1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{\underline{1}}{1}\implies -1}}`

so we're really looking for the equation of a line whose slope is -1 and that it passes through (3 , 2)

`(\stackrel{x_1}{3}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ - 1 \\\\\\ \begin{array} \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}{2}=\stackrel{m}{- 1}(x-\stackrel{x_1}{3}) \\\\\\ y-2=-x+3\implies {\LARGE \begin{array}{llll} y=-x+5 \end{array}}`