=1 ∴ Tangent is given by y−2=1(x−1) y−2=x−1 y=x+1......(i) At B(4,−4) (‌
dy
dx
)(4,−4)=‌
2
−4
=−‌
1
2
∴ Tangent is given by y+4=‌
−1
2
(x−4) 2y+8=−x+4 x+2y+4=0.......(ii) At C(2,2√2) ‌
dy
dx
|(2,2√2)=‌
2
2√2
=‌
1
√2
∴ Tangent is given by y−2√2=‌
1
√2
(x−2) ⇒‌‌√2y−4=x−2 ⇒‌‌x−√2y+2=0..........(iii) Let P,Q,R the vertices of triangle formed by tangents (i), (ii), (iii). On solving Eqs. (i), (ii) and (iii), we get P(−2,−1),Q(−2√2,√2−2) and R(√2,√2+1) Now, ar (∆ABC)=‌