How do I find the most efficient way to find P?

[Poirot1]

Let B=(5,21), C=(40,-14) and $\frac{BP}{PC}$=1/6. What is the most efficent way to find P? My book just states P=(6/7)B+(1/7)C but no idea how they got that.

 

[MarkFL]

You might find this topic helpful:

http://www.mathhelpboards.com/f11/distance-between-two-points-cartesian-plane-2646/

 

[earboth]

Let B=(5,21), C=(40,-14) and $\frac{BP}{PC}$=1/6. What is the most efficent way to find P? My book just states P=(6/7)B+(1/7)C but no idea how they got that.

Draw a sketch.

 

[chisigma]

Let B=(5,21), C=(40,-14) and $\frac{BP}{PC}$=1/6. What is the most efficent way to find P? My book just states P=(6/7)B+(1/7)C but no idea how they got that.

If x and y are the coordinates of P, then it must be...

$\displaystyle f(x,y)= 36\ (x-5)^{2} + 36\ (y-21)^{2} - (x-40)^{2} - (y+14)^{2}=0$ (1)

The (1) is a 'quadratic curve' as illustrated in...

Quadratic Curve -- from Wolfram MathWorld

Kind regards

$\chi$ $\sigma$

 

[chisigma]

If x and y are the coordinates of P, then it must be...

$\displaystyle f(x,y)= 36\ (x-5)^{2} + 36\ (y-21)^{2} - (x-40)^{2} - (y+14)^{2}=0$ (1)

The (1) is a 'quadratic curve' as illustrated in...

Quadratic Curve -- from Wolfram MathWorld

'Monster Wolfram' says that the quadratic curve in this case is a circle...

View attachment 50236 (x-5)^2 + 36 (y-21)^2 - (x-40)^2 - (y+14)^2=0 - Wolfram|Alpha

Kind regards$\chi$ $\sigma$

 

[soroban]

Hello, Poirot!

Did you make a sketch?

Let $B=(5,21),\;C=(40,\text{-}14),\;\dfrac{BP}{PC}=\dfrac{1}{6}$

What is the most efficent way to find $P$ ?

      |
      | (5,21)   +35
      |  B♥ → → → → → → +
      |     *           ↓
      |      Po         ↓
      |         *       ↓ -35
  ----+-----------*-----↓------
      |             *   ↓
      |               * ↓
      |                 ♥C
      |              (40,-14)
      |

Going from $B$ to $C$, we move 35 right and 35 down.

Point $P$ is $\tfrac{1}{7}$ of the way from $B$ to $C$.

The x-coordinate is $\tfrac{1}{7}$ of the way from $5$ to $40$.
. . Hence: .$x \;=\;5 + \tfrac{1}{7}(40-5) \;=\;5 + \tfrac{1}{7}(35) \;=\;5 + 5 \;=\;10$

The y-coordinate is $\tfrac{1}{7}$ of the way from $21$ to $\text{-}14.$
. . Hence: .$y \;=\;21 + \tfrac{1}{7}(\text{-}14 - 21) \;=\;21 + \tfrac{1}{7}(\text{-}35) \;=\;21 - 5 \;=\;16$Therefore, $P$ is at $(10,16).$

 

[CaptainBlack]

Let B=(5,21), C=(40,-14) and $\frac{BP}{PC}$=1/6. What is the most efficent way to find P? My book just states P=(6/7)B+(1/7)C but no idea how they got that.

If your book gives that answer then you have not posted the question as asked, or omitted implied side conditions for the question set.

The quoted answer implies that you want a point P between B and C, while the statement does not so constrain P and defines a locus in the plane.

In future please post the question as asked and include any additional constraints implied by the topic you are studying and or the common conditions in force on a question set.

CB

 

[Poirot1]

I've solved this now thanks to the answers.

 

[earboth]

I've solved this now thanks to the answers.

Even though your problem is solved maybe you want to get some background information. If so have a look here: Circles of Apollonius - Wikipedia, the free encyclopedia