2 x^2-3 x y-2 y^2=0 represents two lines L ₁ and L ₂ . 2 x^2-3 x y-2 y^2-x+7 y-3=0 represents another two lines L ₃ and L ₄ . Let A be the point of intersection of lines L ₁ , L₃ and B be the point of intersection of lines L₂ and L₄ . The area of the triangle formed by lines A B and L₃, L₄ is

Step-by-step solution

Given the equation 2 x^2-3 x y-2 y^2=0 (2 x+y)(x-2 y)=0 Let L ₁: 2 x+y=0, ~L ₂: x-2 y=0 aligned & and 2 x^2-3 x y-2 y^2-x+7 y-3=0 & (2 x+y-1)(x-2 y+3)=0 aligned Let L ₃: x-2 y+3=0, ~L ₄: 2 x+y-1=0 After solving L₁ and L₃ , we get A ( -3 5 , 6 5 ) After solving L ₂ and L ₄ , we get B ( 2 5 , 1 5 ) . After solving L₃ and L₄ , we get C ( -1 5 , 7 5 ) Now, required area = 1 2 | array ccc -3 5 & 6 5 & 1 2 5 & 1 5 & 1 -1 5 & 7 5 & 1 array | aligned & = 1 2 -3 5 ( 1 5 - 7 5 )- 6 5 ( 2 5 + 1 5 )+1 ( 14 25 + 1 25 ) & = 1 2