What is the equation of the line that passes through the pont (6,2) and has a
of -2/3

 y

=

−

2

3

x

+

2

Explanation:

Given that we have the slope and a point on the graph we can use the point slope formula to find the equation of the line.

Point-Slope Formula:  

y

−

y

1

=

m

(

x

−

x

1

)

, where  

m

is the slope of the line and  

x

1

and  

y

1

are x and y coordinates of a given point.

We can summarize the information already given:

m

=

−

2

3

x

1

=

6

y

1

=

−

2

Using this information, we can substitute these values onto the point-slope formula:

y

−

(

−

2

)

=

−

2

3

(

x

−

(

6

)

)

y

+

2

=

−

2

3

(

x

−

6

)

The equation above is the equation of the line in point-slope form. If we wanted to have the equation in  

y

=

m

x

+

b

form then we simply solve the equation above for  

y

y

+

2

=

−

2

3

x

+

12

3

y

+

2

−

2

=

−

2

3

x

+

12

3

−

2

y

=

−

2

3

x

+

12

3

−

2

(

3

3

)

y

=

−

2

3

x

+

12

3

−

6

3

y

=

−

2

3

x

+

6

3

y

=

−

2

3

x

+

2y

=

−

2

3

x

+

2

Explanation:

Given that we have the slope and a point on the graph we can use the point slope formula to find the equation of the line.

Point-Slope Formula:  

y

−

y

1

=

m

(

x

−

x

1

)

, where  

m

is the slope of the line and  

x

1

and  

y

1

are x and y coordinates of a given point.

We can summarize the information already given:

m

=

−

2

3

x

1

=

6

y

1

=

−

2

Using this information, we can substitute these values onto the point-slope formula:

y

−

(

−

2

)

=

−

2

3

(

x

−

(

6

)

)

y

+

2

=

−

2

3

(

x

−

6

)

The equation above is the equation of the line in point-slope form. If we wanted to have the equation in  

y

=

m

x

+

b

form then we simply solve the equation above for  

y

y

+

2

=

−

2

3

x

+

12

3

y

+

2

−

2

=

−

2

3

x

+

12

3

−

2

y

=

−

2

3

x

+

12

3

−

2

(

3

3

)

y

=

−

2

3

x

+

12

3

−

6

3

y

=

−

2

3

x

+

6

3

y

=

−

2

3

x

+

2

Step-by-step explanation:

21.4 i did it already for s1

step-by-step explanation:

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.