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3 years ago

10

if a line with an angle of inclination 120⁰ and passes trough (√3,1),then what is the equation of a line

1 answer:

yawa3891 [41]3 years ago

5 0

Answer:

The equation of a line, having inclination 120° with positive direction of x-axis, .of x-axis, which is at a distance of 3 units from the origin is. 1. See answer ... where α is the angle with the positive X-axis, made by the perpendicular line drawn Now, from equation the equation of the straight line will be.

Step-by-step explanation:

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If two sides of a right triangle are x and y, the hypotenuse, h is:

h^2=x^2+y^2, in this case x and y are 39 and 52 respectively so:

h^2=39^2+52^2

h^2=1521+2704

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Andrei [34K]

Answer:

$\bold{(x+2))^2+(y-2)^2+(z-2)^2=17}$

Step-by-step explanation:

Given the center of sphere is: (-2, 2, 3)

Passes through the origin i.e. (0, 0, 0)

To find:

The equation of the sphere ?

Solution:

First of all, let us have a look at the equation of a sphere:

$(x-a)^2+(y-b)^2+(z-c)^2=r^2$

Where ( $x,y,z$ ) are the points on sphere.

$(a, b, c)$ is the center of the sphere and

$r$ is the radius of the sphere.

Radius of the sphere is nothing but the distance between any point on the sphere and the center.

We are given both the points, so we can use distance formula to find the radius of the given sphere:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Here,

$x_1 =0 \\y_1 =0 \\z_1 =0 \\x_2 =-2 \\y_2 =2 \\z_2 =3$

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$r = \sqrt{(-2-0)^2+(2-0)^2+(3-0)^2}\\\Rightarrow r = \sqrt{4+4+9} = \sqrt{17}$

Therefore the equation of the sphere is:

$(x-(-2))^2+(y-2)^2+(z-2)^2=(\sqrt{17})^2\\\bold{(x+2))^2+(y-2)^2+(z-2)^2=17}$

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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