show that line 3x+4y+5=0 and lines (3x+4y)^2 - 3(4x-3y)^2=0 form an equilateral triangle
Answers
Answer:
The slope of the line 3x+4y+5=0 is m
1
=
4
−3
Let m be the slope of one of the line making an angle of 60
o
with the line 3x+4y+5=0. The angle between the lines having slope m and m
1
is 60
o
∴tan60
o
=
∣
∣
∣
∣
∣
1+m.m
1
m−m
1
∣
∣
∣
∣
∣
, where tan60
o
=
3
∴
3
=
∣
∣
∣
∣
∣
∣
∣
∣
∣
1+m(−
4
3
)
m−(−
4
3
)
∣
∣
∣
∣
∣
∣
∣
∣
∣
∴
3
=
∣
∣
∣
∣
∣
4−3m
4m+3
∣
∣
∣
∣
∣
On squaring both sides, we get,
3=
(4−3m)
2
(4m+3)
2
∴(4−3m)
2
=(4m+3)
2
∴3(16−24m+9m
2
)=16m
2
+24m+9
∴48−72m+27m
2
=16m
2
+24m+9
∴11m
2
−96m+39=0
This is the auxiliary equation of the two lines and their joint equation is obtained by putting m=
x
y
∴ the combined equation of the two lines is
11(
x
y
)
2
−96(y
,
/x)+39=0
∴
x
2
11y
2
−
x
96y
+39=0
∴11y
2
−96xy+39x
2
=0
∴39x
2
−96xy+11y
2
=0 is the joint equation of the two lines through the origin each making an angle of 60
o
with the line 3x+4y+5=0
The equation 39x
2
−96xy+11y
2
=0 can be written as : −39x
2
+96xy−11y
2
=0
i.e. (9x
2
−48x
2
)+(24xy+72xy)+(16y
2
−27y
2
)=0
i.e. (9x
2
+24xy+16y
2
)−3(16x
2
−24xy+9y
2
)=0
i.e (3x+4y)
2
−3(4x−3y)
2
=0
Hence, the line 3x+4y+5=0 and the lines (3x+4y)
2
−3(4x−3y)
2
from the sides of an equilateral triangle.
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