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3 years ago

Which point is on the line that passes through point R and is perpendicular to line PQ? (–6, 10) (–4, –8) (0, –1) (2, 4)

Mathematics

2 answers:

Eddi Din [679]3 years ago

5 0

Answer:

It's B i did the unit test review thing

Step-by-step explanation:

Paraphin [41]3 years ago

4 0

Answer:

(-4, -8)

Step-by-step explanation:

Let the coordinates of common point of the given lines are (x, y),

Thus, the slope of the line passes through the points (a, b) and R(4,2) is,

2 - b

m1 = ------

4 - a

Again, the slope of the line passes through two points P(-6,4) and Q(4,-4),

-4 - 4 -8 -8 -4

m2 = ------ = --------- = --------- = ---------

4- (-6) 4 + 6 10 5

= m1 * m2 = -1

2 - b -4

= ------- x ---- = -1

4 - a 5

8 - 4b

= ----------- = 1

20 - 5a

= 8 - 4b = 20 - 5a

= 5a - 4b = 12 --------- equation 1

For a = -6 and b = -10

5 x -6 - 4 x 10 = -70 ≠ 12

For a = -4 and b = -8

5 x -4 - 4 x - 8 = 12 = 12

For a = 0 and b = -1

5 x 0 - 4 x -1 = 4 ≠ 12

For a = 2 and b = 4

5 x 2 - 4 x 4 = -6 ≠ 12

therefore Second is correct

hope it helps and i get the brainliest.

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If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$