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4 years ago

(5, 3) and (8, -2) Write the standard form of the equation of the line that passes through the given points.

Mathematics

2 answers:

Firlakuza [10]4 years ago

8 0

Answer is given above

Lerok [7]4 years ago

3 0

Slope = (y2 - y1) / (x2 - x1)
(5,3)...x1 = 5 and y1 = 3
(8,-2)..x2 = 8 and y2 = -2
now we sub
slope = (-2 - 3) / (8 - 5) = -5 / 3

y = mx + b
slope(m) = -5/3
use either of ur points..(5,3)...x = 5 and y = 3
now we sub and find b, the y int
3 = -5/3(5) + b
3 = -25/3 + b
3 + 25/3 = b
9/3 + 25/3 = b
34/3 = b

ur equation is y = -5/3x + 34/3...but we need it in standard form

y = -5/3x + 34/3
5/3x + y = 34/3
5x + 3y = 34 <=== standard form

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Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun

WITCHER [35]

Answer:

Required solution (a) $f(x,y,z)=xyz+3z^2+C$ (b) 40.

Step-by-step explanation:

Given,

$F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k$

(a) Let,

$F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}$

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Integrating $f_x$ we get,

$f(x,y,z)=xyz+g(y,z)$

Differentiate this with respect to y we get,

$f_y=xz+g'(y,z)$

compairinfg with $f_y=xz$ of the given function we get,

$g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)$

Then,

$f(x,y,z)=xyz+h(z)$

Again differentiate with respect to z we get,

$f_z=xy+h'(z)=xy+6z$

on compairing we get,

$h'(z)=6z\implies h(z)=3z^2+C$ (By integrating h'(z)) where C is integration constant. Hence,

$f(x,y,z)=xyz+3z^2+C$

(b) Next, to find the itegration,

$\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40$

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3 years ago

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