since function
f(x,y)=cf(x,y)=cf(x,y)=c
c=2y+xx−y−3c=\frac{2y+x}{x-y-3}c=x−y−32y+x
Since this curve passing through (1,2)
substitute x=1 and y=2
then,
c=4+11−2−3c=\frac{4+1}{1-2-3}c=1−2−34+1
c=−54\frac{-5}{4}4−5
so
−54=2y+xx−y−3\frac{-5}{4}=\frac{2y+x}{x-y-3}4−5=x−y−32y+x
-5x+5y+15=8y+4x
9x+3y=15
3x+y=5
y=5-3x
This is the required curve.
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