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, ‌‌
