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