One of the bisectors of the angle between lines
a(x-1)2+2h(x-1)(y-1)+b(y-2)2=0 is
x+2y-5=0
The other bisector is ????
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4 Answers
62(x-1)+2(y-2) = 0
so at the point of intersection, (x-1) = -2(y-2)
hence (4a+b)(y-2)2-4h(y-2)(y-1)=0
Thus, y=2 or (4a+b)(y-1-1)=4h(y-1)
(4a+b-4h)y=4a+b
y = (4a+b)/(4a+b-4h)
but there is a single point of intersection.. so (4a+b)/(4a+b-4h)=2
hence, 4a+b=2(4a+b)-8h
4a+b=8h
now can you solve the rest?
62see there is a single point of intersection fo the bisector and a pair of straight lines
now to make things simple, i have assumed a different form of the line.. x+2y-5=0
that is becasue it makes substitution easier....
once you do that you solve for the 2 roots of y.. but you should have a repeated root.. because there is one point of intersection :)