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Select the equation in slope-intercept form for the line through points (-4,6) and (0,10) and is perpendicular to the line described by y = -x + 2.

A. y=-x -4

B. y=x-10

C. y=x+4

D. y=x+10

User Elimelech
by
8.2k points

2 Answers

5 votes

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+2\qquad \impliedby \begin{array} \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}

now, we know that our line is perpendicular to that one, thus


\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\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 that has a slope of 1 and runs through (0 , 10)


(\stackrel{x_1}{0}~,~\stackrel{y_1}{10})\qquad \qquad \stackrel{slope}{m}\implies 1 \\\\\\ \begin{array}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}{10}=\stackrel{m}{1}(x-\stackrel{x_1}{0}) \\\\\\ y-10=x\implies y=x+10

User Cesar Castro
by
7.5k points
3 votes

Answer:

A I'm pretty sure

Explanation:

hope that helps .-.

User MatthewKremer
by
8.0k points

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