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galben [10]
3 years ago
10

Equation for ( 6, 13) ( 7900, 5 )

Mathematics
1 answer:
tia_tia [17]3 years ago
5 0

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:

y=mx+b

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis

We have the following points through which the line passes:

(x_ {1}, y_ {1}) :( 6,13)\\(x_ {2}, y_ {2}): (7900,5)

We find the slope of the line:

m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {5-13} {7900-6} = \frac {-8} {7894} = - \frac {4} {3947}

Thus, the equation of the line is of the form:

y = - \frac {4} {3947} x + b

We substitute one of the points and find b:

13 = - \frac {4} {3947} (6) + b\\13 = - \frac {24} {3947} + b\\13+ \frac {24} {3947} = b\\b = \frac {51335} {3947}

Finally, the equation is:

y = - \frac {4} {3947} x + \frac {51335} {3947}

Answer:

y = - \frac {4} {3947} x + \frac {51335} {3947}

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Zanzabum

Answer:

\beta=45\degree\:\:or\:\:\beta=135\degree

Step-by-step explanation:

We want to solve \tan \beta \sec \beta=\sqrt{2}, where 0\le \beta \le360\degree.

We rewrite in terms of sine and cosine.

\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}

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Use the Pythagorean identity: \cos^2\beta=1-\sin^2\beta.

\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}

\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)

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\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0

This is a quadratic equation in \sin \beta.

By the quadratic formula, we have:

\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }

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\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }

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When \sin \beta=\frac{\sqrt{2}}{2} , \beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )

\implies \beta=45\degree\:\:or\:\:\beta=135\degree on the interval 0\le \beta \le360\degree.

When  \sin \beta=-\sqrt{2}, \beta is not defined because -1\le \sin \beta \le1

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Hope this helps :)

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