A) \[x-2y+1=0\]
B) \[2x+y-1=0\]
C) \[x+2y-1=0\]
D) \[x-2y-1=0\]
Correct Answer: D
Solution :
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 \[x-2y-1=0\].You need to login to perform this action.
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