Straight Lines and Pair of Straight Lines

2x2+xy−y2−x+2y−1=0
Second degree (homogenous) part of Eq. (i)
2x2+xy−y2=0
assume y=mx, and substitute
2x2+x(mx)−(mx)2=0
Coefficient of x2 should be zero
⇒‌‌2+m−m2=0⇒m=2,−1
m1=2‌ and ‌m2=−1
∵ Both lines are perpendicular their slopes must be
−‌

1

m1

=−‌

1

2

‌ and ‌‌

−1

m2

=1
So, the perpendicular lines has their slopes −1∕2 and 1 and passing through (1,1)
∴ Pair of lines are
‌(y−y1−m1(x−x1))
‌(y−y1−m2(x−x2))=0
‌⇒(y−1+‌

1

9

(x−1))(y−1−(x−1))=0
‌⇒(‌

x

2

+y−‌

3

2

)(y−x)=0
‌⇒‌

1

2

xy−‌

1

2

x2+y2−xy−‌

3

2

y+‌

3

2

x=0
‌⇒−x2−xy+2y2−3y+3x=0
Compare with
‌‌‌ax2+2hxy+by2+2gx+3y=0
‌∴‌‌a=−1,h=+1∕2,b=2g=−3∕2
‌‌ so, ‌‌

b

a

=‌

2

−1

=−2
‌‌ and ‌2(g+h)=2(−‌

3

2

+‌

1

2

)=−2
Hence, ‌

b

a

=2(g+h)