Question

Find the equation of a line perpendicular to y = 4 + x that passes through the point (-3, 3).

Answer 1

Answer:

y=-1/4x+9/4

Step-by-step explanation:

Get slope of the line first (to find slope that is perpendicular, get the negative reciprocal)

4 turns into -1/4

To get the y-intercept, plug in the given coordinate values into this formula:

y=mx+b

3=-3(-1/4)+b

3-3/4=b

b=9/4

b is our y-intercept

9/4 is our y-intercept

Answer 2

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

$y = 4 + x\implies y = \stackrel{\stackrel{m}{\downarrow }}{1}x+4\qquad \impliedby \begin{array}{|c|ll}\cline{1-1}slope-intercept~form\\\cline{1-1}\\y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}}\\\\\cline{1-1}\end{array}$

so therefore

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

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

$(\stackrel{x_1}{-3}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{-1}(x-\stackrel{x_1}{(-3)}) \\\\\\ y-3=-(x+3)\implies y-3 = -x-3\implies y = -x$