Respuesta :
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above
[tex]\bf y = -x-5\implies y = \stackrel{\stackrel{m}{\downarrow }}{-1}x\stackrel{\stackrel{b}{\downarrow }}{-5}\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 it has a slope of -1, or we can say -1/1, thus
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \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 runs through (4,-3)
[tex]\bf P(\stackrel{x_1}{4}~,~\stackrel{y_1}{-3})~\hspace{10em} \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}{4}) \\\\\\ y+3=x-4\implies y=x-7[/tex]
Answer:
I'll give u an example
Step-by-step explanation:
First, put the equation of the line given into slope-intercept form by solving for y. You get y = 2x +5, so the slope is –2. Perpendicular lines have opposite-reciprocal slopes, so the slope of the line we want to find is 1/2. Plugging in the point given into the equation y = 1/2x + b and solving for b, we get b = 6.
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