Question

CAN SOMEONE PLEASE GIVE ME A FULL EXPLANATION ON HOW TO DO THESE TYPES OF PROBLEMS BECAUSE I HAVE ABSOLUTELY NO IDEA ON HOW TO DO THEM AND ALSO PLEASE GIVE THE ANSWER, THANK YOU!!!
Write an equation of the line passing through point P that is perpendicular to the given line.
P(4,−3), y=−x−5

Answer 1

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

$\bf y = -x-5\implies y = \stackrel{\stackrel{m}{\downarrow }}{-1}x\stackrel{\stackrel{b}{\downarrow }}{-5}\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so it has a slope of -1, or we can say -1/1, thus

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}}$

so, we're really looking for the equation of a line whose slope is 1 and runs through (4,-3)

$\bf P(\stackrel{x_1}{4}~,~\stackrel{y_1}{-3})~\hspace{10em} \stackrel{slope}{m}\implies 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{1}(x-\stackrel{x_1}{4}) \\\\\\ y+3=x-4\implies y=x-7$

Answer 2

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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