Question

Find the equations to the straight lines which go through the origin and trisect the portion of the straight line $3x+y=12$ which is intercepted between the axes of coordinates.

Answer 1

$3x+y=12x=0\Rightarrow y=12y=0\Rightarrow x=4$

So the line is interecepted between $A(4,0)$ ans $B(0,12)$

Let the points $P$ and $Q$ trisect the portion of the line $AB$

Now $Q$ divides $AB$ in $2:1$

$\Rightarrow Q(\frac{2\left(0\right)+1\left(4\right)}{2+1},\frac{2\left(12\right)+1\left(0\right)}{2+1})Q(\frac{4}{3},\frac{24}{3})$

So the equation of line joining origin and $Q$ is

$y-0=\left(\frac{\frac{24}{3}-0}{\frac{4}{3}-0}\right)(x-0)y=\frac{24}{4}xy=6x$

Now $P$ divides $AQ$ in $1:1$ , so $P$ is the mid point of $AQ$

$\Rightarrow P(\frac{4+\frac{4}{3}}{2},\frac{0+\frac{24}{3}}{2})P(\frac{8}{3},\frac{12}{3})$

So, the equation of line joining origin to $P$ is

$y-0=\left(\frac{\frac{12}{3}-0}{\frac{8}{3}-0}\right)(x-0)y=\frac{12}{8}x2y=3x$

So, the desired equations are $y=6x$ and $2y=3x$.