Question

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

Answer 1

Answer:

A I'm pretty sure

Step-by-step explanation:

hope that helps .-.

Answer 2

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

$y = \stackrel{\stackrel{m}{\downarrow }}{-1}x+2\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

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)

$(\stackrel{x_1}{0}~,~\stackrel{y_1}{10})\qquad \qquad \stackrel{slope}{m}\implies 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{1}(x-\stackrel{x_1}{0}) \\\\\\ y-10=x\implies y=x+10$