# 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 $-k:l$?

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, $\frac{PA}{PB}=7$.

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 $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{green}{\text{Point P divides the given interval externally in a ratio}\ \ \color{blue}{7:1}}}$$