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brilliants [131]
3 years ago
14

How od you graph slope?

Mathematics
1 answer:
Delicious77 [7]3 years ago
5 0

To graph a slope, first you should get the equation into y-intercept form (y=mx+b) m is the slope and b is the y-intercept. When graphing, a person would vertically (up or down depending whether or not b is positive or negative). Slope is diagonal in most cases. If slope is positive then the line should move up going left to right, and if it is negative the line should move downward. I always remember slope by thinking to get to the elevator then move up or down. An example would be if the equation is y=2x-1. I would start at y-intercept. In this case it would be (0,-1), then I would move right 1 and up 2 because slope is technically 2/1. The new point would be (1, 1). Keep doing the step with slope until you have enough points to draw a line through the points. I hope I was able to help you.
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The equation of the tangent to the circle at A(1,8) is 3x-2y+5=0 and its centre is on the line 7x+3y-1=0 . Find the centre of th
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Step-by-step explanation:

Use the standard form to write two equations using points A and B:

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+

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(

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2

−

h

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2

+

(

0

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2

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(

5

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2

+

(

1

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k

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2

Expand the squares using the pattern

(

a

−

b

)

2

=

a

2

−

2

a

b

+

b

2

4

+

4

h

+

h

2

+

k

2

=

25

−

10

h

+

h

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+

1

−

2

k

+

k

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Combine like terms (noting that the squares cancel):

4

+

4

h

=

25

−

10

h

+

1

−

2

k

Move the k term the left and all other terms to the right:

2

k

=

−

14

h

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22

Divide by 2

k

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−

7

h

+

11

[1]

Evaluate the given line at the center point:

2

h

+

k

−

1

=

0

Write in slope-intercept form

k

=

−

2

h

+

1

[2]

Subtract equation [2] from equation [1]:

k

−

k

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−

7

h

+

2

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11

−

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0

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−

5

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2

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k

=

−

2

(

2

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+

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−

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Substitute the center

(

2

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into the equation of a circle using point A and solve for the value of r:

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−

2

−

2

)

2

+

(

0

−

−

3

)

2

=

r

2

(

−

4

)

2

+

3

2

=

r

2

r

2

=

25

r

=

5

Substitute the center

(

2

,

−

3

)

and #r = 5 into the general equation of a circle, to obtain the specific equation for this circle:

(

x

−

2

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2

+

(

y

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