10(x + 1)2 + (x + 1)( y – 2) – 3(y – 2)2 = 0 …(1)
Put x + 1 = X and y – 2 = Y
∴ (1) becomes
10x2 + xy – 3y2 = 0
10x2 + 6xy – 5xy – 3y2 = 0
2x(5x + 3y) – y(5x + 3y) = 0
(2x – y)(5x + 3y) = 0
5x + 3y = 0 and 2x – y = 0
5x + 3y = 0
5(x + 1) + 3(y – 2) = 0
5x + 5 + 3y – 6 = 0
∴ 5x + 3y – 1 = 0
2x – y = 0
2(x + 1) – (y – 2) = 0
2x + 2 – y + 2 = 0
∴ 2x – y + 4 = 0