Given curve is xy+ax+by=0 Slope of tangent at (1,1) is ‌x⋅‌
dy
dx
+y+a+b⋅‌
dy
dx
=0
‌(‌
dy
dx
)1,1=−[‌
a+y
x+b
](1,1)=‌
−(a+1)
1+b
∴‌‌‌
−(a+1)
1+b
=2 So, ‌2+2b=−a−1 ‌a+2b=−3...(i) ∵(1,1) lies on curve xy+ax+by=0. ∴‌‌a+b=−1....(ii) From Eqs. (i) and (ii), we get ‌b=−2,a=1 ∴‌‌3a+b=3×1+(−2)=1