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3 years ago

Given the two points (-1, 6) and (3, -2), write an equation in point-slope form.

Mathematics

2 answers:

abruzzese [7]3 years ago

8 0

Answer:

y-6 = -2(x+1)

Step-by-step explanation:

First we need to find the slope

m = (y2-y1)/(x2-x1)

= (-2-6)/(3--1)

= (-2-6)/(3+1)

= -8/4

= -2

Then we can point slope form where

y -y1 = m(x-x1)

where m is the slope and x1,y1 is a point

y-6 = -2(x--1)

y-6 = -2(x+1)

Zinaida [17]3 years ago

8 0

Answer:

y - 6 = -2(x + 1)

Step-by-step explanation:

Slope: (-2-6)/(3--1)

= -8/4

= -2

y - 6 = -2(x - -1)

y - 6 = -2(x + 1)

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Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En

ArbitrLikvidat [17]

Answer:

a) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ , b) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ , c) $x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$ , $y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$ .

Step-by-step explanation:

The equation of the circle is:

$x^{2} + (y-1)^{2} = 16$

After some algebraic and trigonometric handling:

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1$

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t$

Where:

$\frac{x}{4} = \cos t$

$\frac{y-1}{4} = \sin t$

Finally,

$x = 4\cdot \cos t$

$y = 1 + 4\cdot \sin t$

a) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ .

b) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ .

c) $x = 4\cdot \cos t''$ , $y = 1 + 4\cdot \sin t''$

Where:

$4\cdot \cos t' = 0$

$1 + 4\cdot \sin t' = 5$

The solution is $t' = \frac{\pi}{2}$

The parametric equations are:

$x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$

$y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$

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