AP EAMCET Engineering 22 Sep 2020 Shift 1 Paper

Let the direction cosines of lines are l1,m1,n1 and l2,m2,n2.
Since, al+bm+cn=0 and, fmn+gnl+hlm=0
So, on eliminating " %, we get
fmn+(gn+hm)(−‌

bm+cn

a

)=0
⇒‌‌afmn=bhm2+cgn2+(bg+ch)mn=0
⇒‌‌bh(‌

m

n

)2+(bg+ch−af)(‌

m

n

)+cg=0 having
roots ‌

m1

n1

and ‌

m2

n2

, so
product of roots =‌

m1m2

n1n2

=‌

ca

bh

Similarly on climinating ' m. we get
gnl+(hl+fni)(−‌

al+cn

b

)=0
⇒‌‌bgnl=ah|2+cfn2+(ch+af)‌ln=0
⇒ah(‌

l

n

)2+(ch+af−bg)(‌

l

n

)+cf=0, having roots ‌

l1

n1

and ‌

l2

n2

,50 product of roots ‌

l1l2

n1n2

=‌

f

ah

∵ The lines are perpendicular.
∴‌‌l1l2+m1m2+n1n2=0