Find the value of $f(P_2)$

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Let $a, b \in\mathbb R$. For $P = (x, y) \in\mathbb R^2$ , define $f(P) = ax + by$. The line $ax + by = 4$ intersects the line segment joining two points $P_1$ and $P_2$ in the plane at a point $R$ such that $P_1R : RP_2 = 1 : 2$. If $f(P_1) = 3$, what is the value of $f(P_2)$?

I was trying this question many times, my teacher gave me this home work. I was trying to solved this problem by using the section formula .But i don't know the further step. If anbody help me , i would be very thankful to him. thank you.

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First we see that $f(R) = 4$ since it lies on line $ax+by =4$.

Second, we see that $f$ is linear, but we need only aditivity. Take any $A(x,y)$ and $B(x',y')$:

$$ f(A+B) = a(x+x')+b(y+y') = ax+by + ax'+by' = f(A)+f(B)$$

Since $\vec{P_2R} = 2\vec{RP_1}$ we have $R-P_2 = 2(P_1-R)$ thus $3R =2P_1+P_2$.

Since $3R =2P_1+P_2$ and $f$ is linear function we have $$f(P_2) = f(3R-2P_1) = 3f(R)-2f(P_1)=3\cdot 4-2\cdot 3 =6$$