Will medal and fan! Use the point slope form, write an equation for the line perpendicular to the line y = -2(x - 5) containing the point (3,4).
Point slope form: \[y-y _{1}=m(x-x _{1})\]
any ideas on what the slope of \(\bf y = -2(x - 5)\) is?
-2 is the slope right?
@jdoe0001
yep that's right :
So what's the next step?
perpendicular slopes are negative reciprocal if slope of first equation is a/b then slope of perpendicular line would be negative b/a
So it would be -1/2?
hmm no.
slope is -2/ a perpendicular line to that one, as Nnesha said, would have a NEGATIVE RECIPROCAL SLOPE, or \(\bf slope=-2\qquad negative\implies +2\qquad reciprocal\implies +\cfrac{1}{2}\)
^
Oh positive 1/2! whoops
So once I find that then what?
so, you're really looking for the equation of a line that has a slope of 1/2 and passes through (3,4) \(\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({\color{red}{ 3}}\quad ,&{\color{blue}{ 4}})\quad \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}=\cfrac{1}{2} \\ \quad \\ % point-slope intercept y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\ \qquad \uparrow\\ \textit{point-slope form}\)
So that would equal y - 4 = 1/2x - 3/2
y = 1/2x +5/2 would be the answer right?
yeap
That's the answer? Sorry I wasn't sure which reply you were saying yes to.
@jdoe0001
Okay thanks!
even more typos
\(\bf y-4=\cfrac{1}{2}(x-3)\implies y=\cfrac{1}{2}x-\cfrac{3}{2}+4\implies y=\cfrac{1}{2}x-\cfrac{3}{2}+\cfrac{8}{2} \\ \quad \\y=\cfrac{1}{2}x+\cfrac{5}{2}\)
Lol it's okay. I got it! :)
:)
o^_^o
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