Respuesta :

mathmate
We need the metric relations (see image for definition of A & B)
which says
AO^2=OB*OP
or
AB^2=OB*BP

We will use the second one, and substitute values
AB=3
OB=4
AB^2=OB*BP => 3^2=4*BP => BP=3^2/4=2.25  
=>
P(4+2.25,0)=(6.25,0)

Using the first relation:
AO=sqrt(4^2+3^2) = 5
OB=4
AO^2=OB*OP => 5^2=4*OP => OP=5^2/4=6.25
=>
x-coordinate of P=6.25



Ver imagen mathmate
gmany
Look at the picture.

The triangle OSR and the triangle RSP are similar therefore:

[tex]\dfrac{|RS|}{|SO|}=\dfrac{|PS|}{|RS|}\\\\|RS|=3;\ |SO|=4;\ |PS|=x[/tex]

substitute

[tex]\dfrac{3}{4}=\dfrac{x}{3}\ \ \ |\text{cross multiply}\\\\4x=9\ \ \ |:4\\\\x=2.25[/tex]
The x-coordinate of point P is equal |OS| + x = 4 + 2.25 = 6.25.

Ver imagen gmany

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