JEE Mains 2020 Model Paper 1 New Pattern

We have a + d + 2√ad = c + b + 2√bc ⇒ a + d = c + b (As ad = bc)
∴ a2+d2+2ad = c2+d2+2ad = c2+b2+2bc ⇒ a2+d2 = c2+b2
∴ (a2+b2)x+b2y+c2 = 0
⇒ (b2+c2)x+b2y+c2 = 0
⇒ b2(y+x)+c2(x+1) = 0
⇒ (y+x)+

c2

b2

(x+1) = 0
Which is of the form L1+λL2 = 0
So, fixed point is (- 1 , 1) = (x0,y0) (Given)
Hence , (x0−1+y0−1) = - 1 + 1 = 0