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Number Theory
Level
pending
Let P(6,0) and Q(12,0) be two fixed points and T(h,k) (where h.k z 0) be a variable point in x-y plane PT and QT meets the y-axis at points R and S respectively and PS meets OT at M (where O is origin). For different values of h and k, the line RM always passes through-
{1,0}
{0,2}
{2,0}
{4,0}
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1 solution
The parametric equation of line PT is given by
p ( t ) = ( 6 , 0 ) + t ( h − 6 , k )
The value of t that gives a zero x-coordinate is t = 6 / ( 6 − h ) , therefore,
R = ( 0 , 6 k / ( 6 − h ) )
Similarly, the parametric equation of line QT is given by
p ( t ) = ( 1 2 , 0 ) + t ( h − 1 2 , k )
The value of t that results in a zero x-coordinate is t = 1 2 / ( 1 2 − h ) , hence,
S = ( 0 , 1 2 k / ( 1 2 − h ) )
Now we have to find the intersection point M between OT and PS. The parametric equations for both are given by
OT: p ( s ) = s ( h , k )
PS: q ( t ) = ( 6 , 0 ) + t ( − 6 , 1 2 k / ( 1 2 − h ) )
Equating p(s) with q(t) , we get the following equations (for the x- and y-coordinates)
s h = 6 − 6 t s k = 1 2 k t / ( 1 2 − h )
Solving for s, by substituting for t (from the first equation) into the second equation.
From the first equation,
t = 6 1 ( 6 − s h )
Substituting this into the second equation,
s k = 2 k ( 6 − s h ) / ( 1 2 − h )
Simplifying and solving for s,
s = 1 2 / ( 1 2 + h )
Therefore, the intersection point is
M = 1 2 / ( 1 2 + h ) ( h , k )
Finally, we get to the line RM, it slope is given by
Slope of RM = ( ( 1 2 k / ( 1 2 + h ) − 6 k / ( 6 − h ) ) / ( 1 2 h / ( 1 2 + h ) ) = 2 ( 6 − h ) − 3 k
Therefore, the equation of RM is
y = 6 − h 6 k − 2 ( 6 − h ) 3 x k
Simplifying and re-arranging, we get,
y = 2 ( 6 − h ) 3 k ( 4 − x )
From this equation, it is evident that the line always passes through the point ( 4 , 0 ) .