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Taya2010 [7]
3 years ago
6

Write an equation of a line in point-slope form that is perpendicular to the line y=1/2x-7 and goes through the point (4,-3)

Mathematics
1 answer:
photoshop1234 [79]3 years ago
5 0

Answer:

y = - 2x + 5

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = \frac{1}{2} x - 7 ← is in slope- intercept form

with slope m = \frac{1}{2}

Given a line with slope m then the slope of a line perpendicular to it is

m_{perpendicular} = - \frac{1}{m} = - \frac{1}{\frac{1}{2} } = - 2, hence

y = - 2x + c ← is the partial equation of the perpendicular line

To find c substitute (4, - 3) into the partial equation

- 3 = - 8 + c ⇒ c = - 3 + 8 = 5

y = - 2x + 5 ← equation of perpendicular line

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The solution is  \frac{\delta  (x,y)}{\delta (u, v)} | = 10e^{-4u}

Step-by-step explanation:

From the question we are told that

        x =  e^{-2a} cos (5v)

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     | \frac{\delta  (x,y)}{\delta (u, v)} | =  | \ det \left[\begin{array}{ccc}{\frac{\delta x}{\delta u} }&{\frac{\delta x}{\delta v} }\\\frac{\delta y}{\delta u}&\frac{\delta y}{\delta v}\end{array}\right] |

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Let \   a =  -2e^{-2u} cos(5v),  \\ b=-2e^{-2u} sin(5v),\\c =-2e^{-2u} sin(5v),\\d=-2e^{-2u} cos(5v)

So

     \frac{\delta  (x,y)}{\delta (u, v)} | = |det  \left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] |

=>    \frac{\delta  (x,y)}{\delta (u, v)} | = | a *  b  - c* d |

substituting for a, b, c,d

=>    \frac{\delta  (x,y)}{\delta (u, v)} | =  | -10 (e^{-2u})^2 cos^2 (5v) - 10 e^{-4u} sin^2(5v)|

=>   \frac{\delta  (x,y)}{\delta (u, v)} | =  | -10 e^{-4u} (cos^2 (5v)   + sin^2 (5v))|

=>  \frac{\delta  (x,y)}{\delta (u, v)} | = 10e^{-4u}

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=========================================================

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