Question

Find the equations of the straight lines which pass through the origin and trisect the portion of the straight line 2x + 3y = 6 which is intercepted between the axes.

Answer 1

Let the line 2x + 3y = 6 intersect the x-axis and the y-axis at A and B, respectively.

At x = 0 we have,
0 + 3y = 6
$\Rightarrow$ y = 2

At y = 0 we have,
2x + 0 = 6
$\Rightarrow$ x = 3

$\therefore A\equiv \left(3,0\right)\mathrm{and}B\equiv \left(0,2\right)$

Let $y=m_{1}x\mathrm{and}y=m_{2}x$ pass through the origin trisecting the line 2x + 3y = 6 at P and Q.
∴ AP = PQ = QB

Let us find the coordinates of P and Q using the section formula.

$$\begin{gathered} P\equiv \left(\frac{2\times 3+1\times 0}{2+1},\frac{2\times 0+1\times 2}{2+1}\right)=\left(2,\frac{2}{3}\right) \\ Q\equiv \left(\frac{1\times 3+2\times 0}{2+1},\frac{1\times 0+2\times 2}{2+1}\right)=\left(1,\frac{4}{3}\right) \end{gathered}$$

Clearly, P and Q lie on $y=m_{1}x\mathrm{and}y=m_{2}x$, respectively.

$$\begin{gathered} \therefore \frac{2}{3}=m_1\times 2\mathrm{and}\frac{4}{3}=m_2\times 1 \\ \Rightarrow m_1=\frac{1}{3}\mathrm{and}{m}_2=\frac{4}{3} \end{gathered}$$

Hence, the required lines are
$y=\frac{1}{3}x\mathrm{and}y=\frac{4}{3}x$
⇒ x − 3y = 0 and 4x − 3y = 0