Question

Find the separate equation of the following pair of straight lines 

(i) 3x² + 2xy – y² = 0 

(ii) 6(x – 1)² + 5(x – 1)(y – 2) – 4(y – 2)² = 0 

(iii) 2x² – xy – 3y² – 6x + 19y – 20 = 0

Answer 1

(i) Factorising 3x² + 2xy – y² we get 

3x² + 3xy – xy – y² = 3x (x + y) – y (x + y) 

= (3 x – y)(x + y) 

So 3x² + 2xy – y² = 0 ⇒ (3x – y) (x + y) = 0 

⇒ 3x – y = 0 and x + y = 0

(ii) 6 (x – 1)² + 5 (x – 1)(y – 2) – 4(y – 2)² = 0 

⇒ 6(x² – 2x +1) + 5(xy – 2x – y + 2) – 4(y² – 4y + 4) = 0 

(i.e) 6x² – 12x + 6 + 5xy – 10x – 5y + 10 – 4y² + 16y – 16 = 0 

(i.e) 6x² + 5xy – 4y² – 22x + 11y = 0 

Factorising 6x² + 5xy – 4y² we get

6x² – 3xy + 8xy – 4y² = 3x (2x – y) + 4y (2x – y) 

= (3x + 4y)(2x – y) 

So, 6x² + 5xy – 4y² – 22x + 11y = (3x + 4y + l )(2x – y + m) 

Equating coefficient of x ⇒ 3m + 21 = -22 ... (1) 

Equating coefficient of y ⇒ 4m – l = 11 ... (2) 

Solving (1) and (2) we get l = -11, m = 0 

So the separate equations are 3x + 4y – 11 = 0 and 2x – y = 0

(iii) 2x² – xy – 3y² – 6x + 19y – 20 = 0 

Factorising 2x² – xy – 3y² we get 

2x² – xy – 3y² = 2x² + 2xy – 3xy – 3y² 

= 2x(x + y) – 3y(x + y) = (2x – 3y) (x + y) 

∴ 2x² – xy – 3y² – 6x + 19y – 20 = (2x – 3y + l)(x + y + m) 

Equating coefficient of x 2m + l = -6 ... (1) 

Equating coefficient of y -3m + l = 19 ... (2) 

Constant term -20 = lm 

Solving (1) and (2) we get l = 4 and m = – 5 where lm = – 20. 

So the separate equations are 2x – 3y + 4 = 0 and x + y – 5 = 0