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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

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)