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3 years ago

The equation of a line parallel to 2x−y+4=0 that passes through the point ( -1, 5) is ___?

Mathematics

2 answers:

soldier1979 [14.2K]3 years ago

5 0

Answer: y = 2x + 7

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

c represents y intercept

m represents the slope of the line.

The equation of the given line is

2x - y + 4 = 0

y = 2x + 4

Comparing with the slope intercept form, slope = 2

If two lines are parallel, it means that they have the same slope. Therefore, the slope of the line passing through (- 1, 5) is 2

To determine the y intercept, we would substitute m = 2, x = - 1 and y = 5 into y = mx + c. It becomes

5 = 2 × - 1 + c

5 = - 2 + c

c = 5 + 2 = 7

The equation becomes

y = 2x + 7

IrinaK [193]3 years ago

3 0

Answer:

y=2x+7

Step-by-step explanation:

2x−y+4=0

y=2x+4

m=2

Line are parallel, so their slope is the same.

A(-1,5)... x1 =-1,y1 =5

y-y1 =m(x-x1)

y1 - 5=2(x-(-1))

y1 - 5=2(x+1)

y1-5=2x+2

y=2x+2+5

y=2x+7

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First find the critical points of f :

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$\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1$

$\dfrac{\partial f}{\partial y}=6y=0\implies y=0$

so the point (1, 0) is the only critical point, at which we have

$f(1,0)=-7$

Next check for critical points along the boundary, which can be found by converting to polar coordinates:

$f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t$

Find the critical points of g :

$\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0$

$\implies\sin t=0\text{ OR }1+5\cos t=0$

$\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi$

where n is any integer. We get 4 critical points in the interval [0, 2π) at

$t=0\implies f(10,0)=155$

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$t=\pi\implies f(-10,0)=235$

$t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299$

So f has a minimum of -7 and a maximum of 299.

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