The correct option is B x2−y2−2x+4y−3=0
x2−y2+2y−1=0
x2−(y2−2y+1)=0
x2−(y−1)2=0
(x−(y−1))(x+y−1)=0
(x−y+1)(x+y−1)=0
Thus the lines are
x−y+1=0
x+y−1=0
Adding both, we get
2x=0
x=0
Substitution in any one equation gives us
y=1.
Thus the point of intersection is (0,1).
Now
L:x+y=3
Since L′ is perpendicular to L
Therefore
L′:x−y=d
It passes through (0,1)
∴d=−1
Hence, L′:x−y+1=0
Thus the combined equation of L and L' is
(x−y+1)(x+y−3)=0
x2−y2−3x+3y+x+y−3=0
x2−y2−2x+4y−3=0