What is the equation of the following line? be sure to scroll down first to see all answer options

The equation for the following graph:

Step-by-step explanation:

- You first need to find the slope by using the slope formula:

(where is the first point and is the second point)

-Use the given points and from the graph for the formula:

Then, you solve:

After you have found the slope, use the slope and the first point for the point-slope formula:

(where is the slope and is the first point)

Then, you solve:

So, the equation for the following graph is .

Cylindrical:

5r^2(sin(t)^2 - cos(t)^2) +r(3cos(t) + 5sin(t)) - 3z^2 + 4z + 12 = 0

Spherical:

5r^2sin(t)^2 (sin(f)^2 - cos(f)^2) - 3r^2 cos(t)^2 + r sin(t) (3cos(f) + 5sin(f)) + 4r cos(t) +12 = 0

Step-by-step explanation:

In cylindrical coordinates:

x = r cos(t)

y = r sin(t)

z = z

Let us reorganize the original equation

−5x^2+3x+5y^2+5y−3z^2+4z+12=0

5 (y^2-x^2) + 3x + 5y - 3z^2 + 4z + 12 = 0

Now, we can replace x and y:

5 (r^2 sin(t)^2 - r^2 cos(t)^2) + 3rcos(t) + 5r sin(t) - 3z^2 + 4z + 12 = 0

5r^2(sin(t)^2 - cos(t)^2) +r(3cos(t) + 5sin(t)) - 3z^2 + 4z + 12 = 0

In spherical coordinates:

x = r sin(t) cos(f)

y = r sin(t) sin(f)

z = r cos(t)

Let us reorganize the equation:

5 (y^2-x^2) - 3z^2  + 3x + 5y + 4z + 12 = 0

5r^2sin(t)^2 (sin(f)^2 - cos(f)^2) - 3r^2 cos(t)^2 + r sin(t) (3cos(f) + 5sin(f)) + 4r cos(t) +12 = 0