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I need to calculate the coordinate of Point P3 given the following data.

Distance D from P1 = 3
P1 coord = (1,3)
P2 coord = (10,6)

please help me to solve this issue..

i understand I have to use the distance formula between to coordinate.. but my math is very rusty can find the way to calculate the inverse formula.

$$D(P1-P3) = \sqrt{(P2x - P1x)^2 + (P2y - P1y)^2}$$

but... I don't know how to procede.. I need to calculate X,Y of point P3.

can someone show me what the step to take in order to get the coordinate of P3?

Thanks a lot

pic

Damiano Miazzi
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    See this. – John Douma Apr 12 '22 at 05:38
  • Start with determining vector $\vec{P_1P_2}=(,)$. Then find length $P_1P_2=\sqrt{}$. Vector $\vec{P_1P_3}=\vec{P_1P_2}\frac{P_1P_3}{P_1P_2}=(,)$ Then $P_3=(,*)$. – Ivan Kaznacheyeu Apr 12 '22 at 15:21
  • Formula in question is not correct. Correct writing is $P_1P_3=\sqrt{(P_{1x}-P_{3x})^2+(P_{1y}-P_{3y})^2}$. If one can use only this formula, one can find $P_1P_2$, then $P_2P_3=P_1P_2-P_1P_3$, then one can solve simultaneous equation for $P_1P_3$ and $P_2P_3$ to find $P_{3x}$ and $P_{3y}$. But this method is weird. – Ivan Kaznacheyeu Apr 12 '22 at 15:29

1 Answers1

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  1. Use "2-point form" to get L, the equation of P1P2.

---> $L: \dfrac {y - 2}{x - 1} = \dfrac {6 - 2}{10 - 1}$

  1. Form C, the equation of the circle (centered at P1 with radius = 3). $C: (x - 1)^2 + (y - 2)^2 = 3^3$

  2. P3 is located at the intersection of L and C. Therefore, solve L and C to get P3. [Will get 2 sets of answers from the combined quadratic equation. Choose the one between P1 and P3.]

Mick
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