There are two points, A and B.

The point A is (1,1) and point B is (5,6)

Point P is along the line segment of AB, and it makes a ratio of AP to BP to 1:2.

What is point P?

Round to tenth decimal

Please tell me what formula you used to get this, really confused.

Guest Feb 12, 2019

#1**+2 **

Point p is 1/3 of the way from AP to BP this will nake the ratio AP:BP 1:2 (1+2) =3

Let's find x coordinate first from 1 to 5 is 4 units we need 1/3 4 x 1/3 = 4/3 add this to A

1 + 4 /3 = 7/3 = x

Now for y = 1 to 6 is 5 5 x 1/3 = 5/3 add this to A

1+5/3 = 8/3 = y

7/3, 8/3 = p = (2.3,2.7) rounded

ElectricPavlov Feb 12, 2019

#1**+2 **

Best Answer

Point p is 1/3 of the way from AP to BP this will nake the ratio AP:BP 1:2 (1+2) =3

Let's find x coordinate first from 1 to 5 is 4 units we need 1/3 4 x 1/3 = 4/3 add this to A

1 + 4 /3 = 7/3 = x

Now for y = 1 to 6 is 5 5 x 1/3 = 5/3 add this to A

1+5/3 = 8/3 = y

7/3, 8/3 = p = (2.3,2.7) rounded

ElectricPavlov Feb 12, 2019

#2**+5 **

**There are two points, A and B.**

**The point A is (1,1) and point B is (5,6)**

**Point P is along the line segment of AB, and it makes a ratio of AP to BP to 1:2. **

**What is point P?**

**Round to tenth decimal**

\(\begin{array}{|rcll|} \hline \vec{P} &=& \vec{A}+ \lambda \left( \vec{B}-\vec{A} \right) \quad | \quad \lambda = \dfrac{1}{1+2} = \dfrac{1}{3} \\\\ \vec{P} &=& \vec{A}+ \dfrac{1}{3}\left( \vec{B}-\vec{A} \right) \quad | \quad \vec{A}=\dbinom{1}{1},\ \vec{B}=\dbinom{5}{6} \\\\ \vec{P} &=& \dbinom{1}{1}+ \dfrac{1}{3}\left( \dbinom{5}{6}-\dbinom{1}{1} \right) \\\\ \vec{P} &=& \dbinom{1}{1}+ \dfrac{1}{3} \dbinom{5-1}{6-1} \\\\ \vec{P} &=& \dbinom{1}{1}+ \dfrac{1}{3} \dbinom{4}{5} \\\\ \vec{P} &=& \dbinom{1+\frac{4}{3}}{1+\frac{5}{3}} \\\\ \vec{P} &=& \dbinom{ \frac{7}{3}}{ \frac{8}{3}} \\\\ \mathbf{\vec{P}} & \mathbf{=} & \mathbf{\dbinom{2.3}{2.7}} \\ \hline \end{array}\)

**P = (2.3, 2.7)**

heureka Feb 13, 2019