For
△ABC, we have sides are
2x2−y2=0,x+y−1=0 (√2x−y)(√2x+y)=0,x+y−1=0
√2x−y=0,√2x+y=0,x+y−1=0
On solving these, we get vertices
(0,0),(√2−1,2−√2),(−√2−1,2+√2)
For
∆PQR, we have sides are
2x2−5xy+2y2=0,7x−2y−12=0
(2x−y)(x−2y)=0,7x−2y−12=0
2x−y=0,x−2y=0,7x−2y−12=0
On solving these, we get vertices
(0,0),(4,8),(2,1) Now, centroid of
△ABC =(,) Orthocentre of
△PQR,
∴ Required distance
=GH =√100+16=√116=2√29