Complete the point slope equation of the line through (-5,7) and (-4,0)y-7=

2 answers:

7 0

$\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{7})\qquad
(\stackrel{x_2}{-4}~,~\stackrel{y_2}{0})
\\\\\\
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-7}{-4-(-5)}\implies \cfrac{0-7}{-4+5}\implies \cfrac{-7}{1}\implies -7
\\\\\\
\begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-7=-7[x-(-5)]\implies y-7=-7(x+5)$

deff fn [24]3 years ago

3 0

Point slope form: y-7= -7(x+5)

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$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
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-------------------------------\\\\$

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\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
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if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
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\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

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