Math, asked by rahulwaradkar452, 9 months ago

Find slope of line L3 : 3x – y + 6 = 0

Answers

Answered by sara6857
1

Answer:

L

1

≡3x−2y−6=0

L

2

≡3x+4y+12=0

L

3

≡3x−8y+12=0

Simultaneously solving the equations L

1

and L

2

gives one vertex (A) as (0,−3).

SImilarly, solving the equations L

2

and L

3

gives second vertex (B) as (−4,0).

Finally, solving the equations L

3

and L

1

gives third vertex (C) as (4,3).

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Circumcentre is point of interesection of perpendicular bisectors of all the three points.

Let us first find the perpendicular bisector of AB:

Mid-point of AB≡(

2

0−4

,

2

−3+0

)≡(−2,−

2

3

)

Slope of AB=

x

2

−x

1

y

2

−y

1

=

−4−0

0−(−3)

=−

4

3

Therefore, slope of line perpenicular to AB=

3

4

(Since product of slopes of perpendicular lines = -1)

Therefore, equation of bisector:

(y+

2

3

)=

3

4

(x+2)⟹8x−6y+7=0 (1)

(1) is the required perpendicular bisector.

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Now, finding the perpendicular bisector of BC

Mid-point of BC=(0,

2

3

)

Slope of BC=

8

3

Therefore, slope of line perpenicular to AB=−

3

8

Therefore, equation of bisector:

y−

2

3

=−

3

8

(x−0)⟹16x+6y−9=0 (2)

(2) is the required perpendicular bisector.

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Simultaneously solving (1) and (2) gives:

x=

12

1

,y=

18

23

∴β=

18

23

,α=

12

1

⟹3(β−α)=3(

18

23

12

1

)=3(

36

46−3

)=

12

43

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