Find the value of k, if the equation 2x² + kxy - 6y² + 3x + y +1=0 represents a pair of straight lines.
Find the point of intersection of the lines and the angle between the straight lines for this value of k.
Answers
Answer:
12x
2
−10xy+2y
2
+11x−5y+λ=0
For pair of straight line
ax
2
+2hxy+by
2
+2gx+2fy+c=0
⎣
⎢
⎢
⎡
a
h
g
h
b
f
g
f
c
⎦
⎥
⎥
⎤
=0
Here, $$a=12,h=\cfrac=-{10}{2}=-5,b=2,g=\cfrac{11}{2}
,\cfrac=-{5}{2}& c=\lambda$$
⎣
⎢
⎢
⎡
12
−5
2
11
−5
2
−
2
5
2
11
−
2
5
λ
⎦
⎥
⎥
⎤
=0
=12(12λ−
4
25
)+5(−5λ+
4
55
)+
2
11
(
2
25
−11)=0
or,12(8λ−25)−100λ+275+33=0
96λ−300−100λ+308=0
or,−4x+8=0
λ=2
Equation is
12x
2
−10xy+2y
2
+11x−5y+2=0
12x
2
−4xy+8x+2y
2
−6xy−4y+3x−y+2=0
4x(3x−y+2)−2y(−y+3x+2)+(4x−2y+1)=0
Two eqations are
3x−y+2=0×2→(1)
4x−2y+1=0×1→(2)
6x−2y+4=0
4x−2y+1=0
−−−−−−−−
2x+3=0
x=−
2
3
Putting value of x in (2)
4(−
2
3
)−2y+1=0
−6−2y=1=0
y=−
2
5
So point of intersection is
(−
2
3
,
2
5
)
y=3x=2 herem
1
=3
m
1
=3
y=
2
4x+1
&m
2
=
2
4
=2
So, angle between them
=tan
−1
(±
1+m
1
m
2
m
1
−m
2
)
=tan
−1
(
1+6
3−2
)&tan
−1
(−
1+6
3−2
)
=tan
−1
(
7
1
)&tan
−1
(−
7
1
)
and other angls are
=π−tan
−1
(
7
1
)&π−tan
−1
(−
7
1
)
Pair of angle bisector
a
1
2
+b
1
2
a
1
x+b
1
y+c
1
=±
a
2
2
+b
2
2
a
2
x+b
2
y+c
2
where equations are a
1
x+b
1
y+c
1
=0 & a
2
x+b
2
y+c
2
=0
So,
3
2
+1
2
3x−y+2
=±
4
2
+2
2
4x−2y+1
10
3x−y+2
=±
20
4x−2y+1
or, 3x−y+2=±
2
4x−2y+1
Answer:
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STEP BY STEP EXPLANATION
