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# coordinates

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In the coordinate plane, A = (0,0), B = (0,4), and C = (5,0).  If P is a point inside triangle ABC such that AP = sqrt(10) and BP = 3*sqrt(2), then find CP.

Dec 7, 2020

#1
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Here's one  method  to  find P

Construct  a  circle   with a center at  (0,0)  with a radius  of   sqrt (10)

The equation of this circle  is  x^2  +  y ^2 =  10     (1)

Similarly.....at B  construct  a circle  with a radius  of  3sqrt (2)   =   sqrt (18)

The equation of this circle  is  x^2 + ( y-4)^2  = 18     (2)

Subtracting (1)  from (2)   we  have that

( y - 4)^2  - y^2  =  8   simplify

y^2  - 8y  +  16  - y^2   =   8

-8y   +  16  =  8

-8y =  -8      →    y = 1

And to  find  x  we have that

x^2  + (1)^2    = 10

x^2   +  1   =10

x^2  = 9       take the  positive  root

x   =  3

So  P  =  ( 3,1)

Using the  distance  formula

CP  = sqrt  [( 5-3)^2  +  ( 1 - 0)^2  ]  =    sqrt [ 2^2  + 1^2 ]  =     sqrt (5)   Dec 7, 2020
#2
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In the coordinate plane, A = (0,0), B = (0,4), and C = (5,0).  If P is a point inside triangle ABC such that AP = sqrt(10) and BP = 3*sqrt(2), then find CP.

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Use the law of cosines to find ∠BAP.    Find ∠PAC

Using ∠PAC and line AP, calculate line segments AD and PD

CP = sqrt[PD2 + (AC - √8)2]     ==>   CP ≈ 2.591 Dec 8, 2020
edited by Dragan  Dec 8, 2020
#3
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I made a mistake!!! I used the wrong angle. Instead of √2 and √8  should be  1 and 3

Please don't trust me. I'm not Nikola Tesla. 