factorise 4+x²+9y²+4x-6xy-12y
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Answer:
Let lx+my+c
1
=0 & lx+my+c
2
=0
be two parallel lines represented by the given equation then we have
ax
2
+2hxy+by
2
+2gx+2fy+c=(lx+my+c
1
)(lx+my+c
2
)
On comparing
a=l
2
,b=m
2
,2h=2lm,l(c
1
+c
2
)=2g
and m(c
1
+c
2
)=2f,c
1
c
2
=c
Now distance between the parallel lines
=
l
2
+m
2
∣c
1
−c
2
∣
=
l
2
+m
2
(c
1
+c
2
)
2
−4c
1
c
2
=
a+b
(
l
2g
)
2
−4c
=
a+b
a
4g
2
−4c
=2
a(a+b)
g
2
−ac
i. e. Reason (R) is false
Again x
2
+6xy+9y
2
+4x+12y−5=0
are two parallel lines given by (x+3y−1)(x+3y+5)=0
i.e. x+3y−1=0 & x+3y+5=0
and distance between them
=2
a(a+b)
g
2
−ac
=2
1(1+9)
4+5
=
10
6
so the Assenion (A) is true As Assertion (A) is true & Reason (R) is false,
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