Vector algebra triangle problem

Question

Let ABC be a triangle and M and N are given points such that $BM:MC=1:4$ and $CN:NA=2:5$. Let P be the intersection point of $\vec{AM}$ and $\vec{BN}$. Write $\vec{AP}$ as linear combination of vectors $a=\vec{BC}$ and $c=\vec{AB}$

Answer 1

2 solutions:

A) Using analytic geometry

The property you want to establish is invariant by an affine transform ; therefore, you can assume WLOG that $A(0,7),B(-5,0),C(0,0)$ ; a consequence is that the coordinates of points $M$ and $N$ are $M(-4,0)$ and $N(0,2)$.

It is easy to establish that straight lines AM (resp. BN) have equations $y=\tfrac{7}{4}x+7$ (resp. $y=\tfrac{2}{5}x+2$.

Thus intersection point $P$ has coordinates $P(- \ \tfrac{100}{27}; \tfrac{14}{27})$

We now have to find $\lambda$ and $\mu$ such that:

$$\lambda \binom{5}{0}+\mu \binom{-5}{-7}=\binom{- \ \tfrac{100}{27}-0}{\ \ \ \ \tfrac{14}{27}-7}$$

finally giving

$$\tag{*}\lambda = \dfrac{5}{27} \ \ ; \ \ \mu = \dfrac{25}{27} $$

B) Using barycentrical coordinates

It is not difficult to establish that barycentrical coordinates of $P$ are $(\tfrac{2}{27},\tfrac{20}{27},\tfrac{5}{27})$.

Thus we can write

$$\tag{1}P=\tfrac{2}{27}A+\tfrac{20}{27}B+\tfrac{5}{27}C$$

Besides, relationship :

$$\tag{2}\vec{AP}=\lambda\vec{BC}+\mu\vec{AB}$$

can be written under the form:

$$P-A=\lambda (C-B)+\mu (B-A) \ \ \iff$$

$$\tag{3}P=(1-\lambda)A+(\mu-\lambda)B+\lambda C.$$

Comparing (1) and (3), we get the proportionality relationships:

$$\dfrac{1-\lambda}{2}=\dfrac{\mu-\lambda}{20}=\dfrac{\lambda}{5}$$

out of which one finds for $\lambda$ and $\mu$ the same values as in (*).