TS EAMCET 4 May 2018 Shift 1 Solved Paper

The given relation is,
al+bm+cn=0
and
mn+nl+lm=0
Thus,
(

al+bm

−c

)(1+m)+lm=0
al2+(a+b−c)lm+bm2=0
a (

l

m

)2+(a+b−c)(

l

m

)+b=0 ……. (I)
Let the roots of the equation be

l1

m1

and

l3

m2

Therefore,

l1l2

m1m2

=

b

a

Thus,

l1l2

n1n2

=

c

a

Hence,

l1l2

1 a

=

m1m2

1 b

=

n1n2

1 c

The lines whose direction cosine are perpendicular if,

1

a

+

1

b

+

1

c

=0