AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Solve the lines y=mx and x−y+10=0 for coordinate ofpoint A and B.
A(

10

m−1

,

10m

m−1

),B(

20

m−1

,

20m

m−1

)
The distance OA=√

100

(m−1)2

+

100m2

(m−1)2

=

10√1+m2

(m−1)

The distance OB=√

400

(m−1)2

+

400m2

(m−1)2

=

20√1+m2

(m−1)

Let a point P(h,k) lies on the line y=mx
k=mh⇒m=

k

h

The distance OP=√h2+k2
Now, OA, OP, OB are in HP.
Therefore,

1

OA

,

1

OP

,

1

OB

in AP.

1

OP

=

1

OA

+

1

OB

2

√h2+k2

=

m−1

10√1+m2

+

m−1

20√1+m2

2

√h2+k2

=

m−1

10√1+m2

[1+

1

2

]

2

√h2+k2

=

( k h −1)

10√1+( k h )2× 3 2

2

√h2+k2

=

3(k−h)

20√h2+k2

3(k−h)=40
Therefore, the locus of the point is
3x−3y+40=0