TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

For △ABC, we have sides are
2x2−y2=0,x+y−1=0
On solving these, we get vertices
For ∆PQR, we have sides are
On solving these, we get vertices (0,0),(4,8),(2,1)
Now, centroid of △ABC
=(‌

−2

3

,‌

4

3

)
Orthocentre of △PQR,
∴ Required distance =GH
=√100+16=√116=2√29