Paragraph: Let ABC be a triangle equation of whose sides are AB x +2 y =3, AC 2 x + y =3 , and BC x + y =4 . Circles S ₁, ~S ₂ and S ₃ are drawn taking AB , AC and BC as diameter respectively. Now L ₁₂=0, ~L ₃₁=0 and L ₂₃=0 are the radical axis of S ₁ and S ₂, ~S ₃ and S ₁ , and S ₂ and S ₃ respectively. These radical axis intersects the sides BC , AC and AB at points D , E and F respectively. On the basis of above i

Step-by-step solution

Coordinates of A , B and C are A (1,1) , B (5,-1), C (-1,5) . Now equation of the circles taking AB , AC and BC as diameter respectively will be aligned S₁ & (x-1)(x-5)+(y-1)(y+1)=0 & x^2+y^2-6 x+4=0 aligned aligned S ₂ & ( x -1)( x +1)+( y -1)( y -5)=0 & x ^2+ y ^2-6 y+4=0 ~S ₃ & (x-5)(x+1)+(y+1)(y-5)=0 & x^2+y^2-4 x-4 y-10=0 aligned Equation of radical axis of S₂ and S₃ will be aligned L ₂₃ & 4 x -2 y +14=0 & 2 x - y +7=0 aligned clearly radical centre will be the point of intersection of L₁₂=0 & L₂₃=0 i.e., (-7,