• # question_answer One bisector of the angle between the lines given by $a{{(x-1)}^{2}}+2h\,(x-1)y+b{{y}^{2}}=0$ is $2x+y-2=0$. The other bisector is A)            $x-2y+1=0$                         B)            $2x+y-1=0$ C)            $x+2y-1=0$                         D)            $x-2y-1=0$

We have $a{{(x-1)}^{2}}+2h(x-1)y+b{{y}^{2}}=0$            or $a{{(x-1)}^{2}}+2h(x-1)(y-0)+b{{(y-0)}^{2}}=0$            This equation represents a pair of straight lines intersecting at (1, 0). Therefore shifting the origin at    (1, 0), we have $x=X+1,\,y=Y+0$ and the equation reduces to $a{{X}^{2}}+2hXY+b{{Y}^{2}}=0$                            .....(i)            One of the bisectors of the angles between the lines given by (i) is $2x+y-2=0$or $2(X+1)+Y-2=0$ i.e. $2X+Y=0$. Since the bisector are always at right angle, therefore the other bisector is $X-2Y=0$            i.e., $x-1-2y=0$or $x-2y-1=0$.