Linear Functions: Division of an interval The lines $y= 4x + 2$ and $x+2y=6$ intersect at point $P$. 
i) Find the coordinates of point $P$.
$$y= 4x + 2\ldots(1)$$
$$x+2y=6 \ldots(2)$$
Sub ($1$) into ($2$)
$$x+2(4x + 2) =6$$
$$9x=2$$
$$x=2/9$$
sub $x = 2/9$ into ($1$) to find $y$
$$y=26/9$$
Therefore the coordinates of $P$ is $(\frac{2}{9}, \frac{26}{9})$.
ii) Find the ratio in which $P$ divides the interval $(1,\frac{33}{9})$ and $(\frac{1}{3},3)$
Using the interval division formula, and letting the ratio be $k:l$, I got two equations, $\frac{k (1/3) + l(1) }{k+l}$ and $\frac{k (3) + l(33/9) }{k+l}$.
However when I simplified both, I ended up getting $7l+k=0$ on both sides, which left me stuck as I was not sure where I went wrong. Should I try with the ratio 
$-k:l$?
 A: HINT:
The distance between two points $(x_1,y_1), \; (x_2,y_2)$ is:
$\sqrt{(x_1-x_2)^2+ (y_1-y_2)^2}$.
In your case, let $A(1,\frac{33}{9}),\; B(\frac{1}{3},3)$. 
Then $PA=\frac{7}{9}\sqrt 2$ and $PB=\frac{1}{9}\sqrt 2$.
Thus, $\frac{PA}{PB}=7$.
A: Notice,let $m:n$ be the ration in which the given point $P\left(\frac{2}{9}, \frac{26}{9}\right)$ divides the line joining the points $\left(1, \frac{33}{9}\right)$   & $\left(\frac{1}{3}, 3\right)$ respectively  
then the coordinates of the point P are calculated using division formula as follows $$P\equiv \color{red}{\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)}$$ Now, substituting the corresponding values we get 
$$\left(\frac{2}{9}, \frac{26}{9}\right)\equiv \left(\frac{\frac{1}{3}\cdot m+1\cdot n}{m+n}, \frac{3\cdot m+n\cdot \frac{33}{9}}{m+n}\right)$$
Now, comparing the corresponding $x$ coordinates, we get $$\frac{\frac{1}{3}\cdot m+1\cdot n}{m+n}=\frac{2}{9}\iff m+7n=0$$  $$\frac{m}{n}=-\frac{7}{1}$$
Similarly, comparing the y-coordinate, we get
$$\frac{3\cdot m+n\cdot \frac{33}{9}}{m+n}=\frac{26}{9}\iff m+7n=0$$
 $$\frac{m}{n}=-\frac{7}{1}$$
Negative sign indicates that the point $P$ divides the interval $\left(1, \frac{33}{9}\right)$ & $\left(\frac{1}{3}, 3\right)$ externally in a ratio $7:1$
Thus, we conclude that 
 $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{green}{\text{Point P divides the given interval externally in a ratio}\ \ \color{blue}{7:1}}}$$
