Given that, √(x−3)2+(y−1)2+√(x+3)2+(y−1)2=6 ⇒√(x−3)2+(y−1)2=6−√(x+3)2+(y−1)2 On squaring both sides, we get (x−3)2+(y−1)2=36+(x+3)2+(y−1)2−12√(x+3)2+(y−1)2 ⇒x2−6x+9+(y2−2y+1)=36+x2+6x+9+y2−2y+1−12√(x+3)2+(y−1)2 ⇒(x+3)=√(x+3)2+(y−1)2 On squaring both sides of above equation ⇒(x+3)2=[(x+3)2+(y−1)2] ⇒(y−1)2=0 ⇒ Which represents a pair of straight lines.