Q.The pair of lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ and $a'x^2 + 2h'xy + b'y^2 + 2g'x + 2f'y + c' = 0$ have the same angle bisectors if

\dfrac{a - b}{h} = \dfrac{a' - b'}{h'}

\dfrac{a + b}{h} = \dfrac{a' + b'}{h'}

a + b = a' + b'

a - b = a' - b'

The correct solution involves applying the fundamental concept to derive the final value step by step...

Already a member? Login here

Q.The pair of lines ax2+2hxy+by2+2gx+2fy+c=0ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0ax2+2hxy+by2+2gx+2fy+c=0 and a′x2+2h′xy+b′y2+2g′x+2f′y+c′=0a'x^2 + 2h'xy + b'y^2 + 2g'x + 2f'y + c' = 0a′x2+2h′xy+b′y2+2g′x+2f′y+c′=0 have the same angle bisectors if