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klio [65]
3 years ago
9

A line passes through a point (2,8) and has a slope of -5. Write an equation in point-slope form for this line.

Mathematics
1 answer:
vovangra [49]3 years ago
8 0
8= -5(x-2) or -5x-y+18
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\large \boxed{sin \theta = tan \theta cos \theta} \\  \large \boxed{tan^{2}  \theta + 1 =  {sec}^{2} \theta}

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\large{ {4}^{2}  + 1 =  {sec}^{2} \theta } \\  \large{16 + 1 =  {sec}^{2} \theta } \\  \large{ {sec}^{2}  \theta = 17}

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\large \boxed{sec \theta =   \frac{1}{cos \theta} } \\  \large \boxed{ {sec}^{2}  \theta =  \frac{1}{ {cos}^{2}  \theta} }

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\large{  {cos}^{2}  \theta =  \frac{1}{17}}   \\ \large{cos \theta =  \frac{ \sqrt{1} }{ \sqrt{17} } } \\  \large{cos \theta =  \frac{1}{ \sqrt{17} }  \longrightarrow  \frac{ \sqrt{17} }{17} }

Since the given domain is 180° < θ < 360°. Thus, the cosθ < 0.

\large{cos \theta =   \cancel\frac{ \sqrt{17} }{17} \longrightarrow cos \theta =  -  \frac{ \sqrt{17} }{17}}

Then use the Identity of sinθ = tanθcosθ to find the sinθ.

\large{sin \theta = 4 \times ( -  \frac{ \sqrt{17} }{17}) } \\  \large{sin \theta =  -  \frac{4 \sqrt{17} }{17} }

Answer

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