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Sagot :
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
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above
[tex]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}[/tex]
so therefore
[tex]\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}}[/tex]
so we're really looking for the equation of a line whose slope is -1 and passes through (-3 , 3)
[tex](\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[/tex]
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