The lines represented b the equations 23 x^2-48 x y+3 y^2=0 and 2 x+3 y+4=0 form

Step-by-step solution

(c) We have, aligned & 23 x^2-48 x y+3 y^2=0 & 3 y^2-48 x y+23 x^2=0 & Here, M₁+M₂=16 & and M₁ M₂= 23 3 & m₁-m₂= (m₁+m₂ )^2-4 (m₁ m₂ ) & = (16)^2-4 23 3 = 256- 92 3 = 768-92 3 = 26 3 & m₁=8+ 13 3 and m₂=8- 13 3 & = | m₁-m₂ 1+m₁ m₂ |= | 26 3 1+ 23 3 |= 3 & aligned We know that, than 60^ So, =60^ Slope of line 2 x+3 y+4=0 is - 2 3 Angle between line of slope 8+ 13 3 and - 2 3 is aligned & = 8+ 13 3 + 2 3 1+ (8+ 13 3 ) ( 2 3 ) = 3 & =60^ & =60^ aligned Hence, line from a equilateral triangle.