Write the equation of the line that passes through (−1, 5) and has a slope of 3 in point-slope form. (2 points)

Mathematics

1 answer:

Sonbull [250]4 years ago

8 0

-1 = 3(5) + b
-1 = 15 + b
-16 = b

y = 3x -16
y - 5 = 3(x +1)

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OlgaM077 [116]

(5-12)^2+8/(-2)
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8 0

4 years ago

Find parametric equations which describe each curve. No particular orientation is required.

MA_775_DIABLO [31]

Answer:

The equation of the line is $(x(t),y(t))=(1-4t,2+6t)$ and the equation of the circle is $F(t)= (3cos(t)+4,3sin(t)+1)$ .

Step-by-step explanation:

(a) Given: The given points are $(1,2)$ and $(-3,8)$ .

To find: The parametric equation of line containing points $(1,2)$ and $(-3,8)$ .

We know that the parametric equation of line containing $(x_{1} ,y_{1} )$ and $(x_{2} ,y_{2} )$ is given by $(x(t),y(t))=(x_{1}+(x_{2}-x_{1})t,y_{1}+(y_{2}-y_{1})t)$ where $t$ ∈ $[0,1]$ .

Now, $x(t)=1+(-3-1)t$

i.e, $x(t)=1-4t$

And, $y(t)=2+(8-2)t$

i.e, $y(t)=2+6t$

Hence, the required parametric equation of the line is $(x(t),y(t))=(1-4t,2+6t)$ .

(b) Given: The radius of circle is 3 and centre is $(4,1)$ .

To find: The parametric equation of circle with radius 3 and centre $(4,1)$ .

We know that parametric equation of circle with radius $r$ and centre $(h,k)$ is given by $F(t)= (x(t),y(t))$ where $x(t)=rcos(t)+h$ and $y(t)=rsin(t)+k$ .