TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

Equation of line are x+y−1=0 and x−y+1=0 Solving equation we get, (0,1)
∴R(0,1)
Perpendicular distance from P to line x+y−1=0
and x−y+1=0 are ‌

1

√2

and ‌

1

√2

respectively
∴‌‌PS=PQ and △PSR and ∆PQR are congruent triangle
∴ Area of quadrilateral PQRS=PQ×PS
=‌

1

√2

×‌

1

√2

=‌

1

2