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

$\text{Let the line}2x+3y=6\text{intersect the x- axis and the y-axis at A and B, respectively,}$

$Atx=0\text{we have,}$

$0+3y=6$

$\Rightarrow y=2$

$Aty=0\text{we have,}$

$2x+0=6$

$\Rightarrow x=3$

$\therefore A=(3,0)andB=\left(02\right)$

$Lety=m_{1}xandy=m_{2}x\text{pass through the origin trisecting the line}2x+3y=6atPandQ$

$\therefore AP=PQ=QB$

$\text{Let us find the coordinates of P and Q using the section formula.}$

$P=(\frac{2\times 3+1\times 0}{2+1},\frac{2\times 0+1\times 2}{2+1})=(2,\frac{2}{3})$

$Q=(\frac{1\times 3+2\times 0}{2+1},\frac{1\times 0+2\times 2}{2+1})=(1,\frac{4}{3})$

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

$\therefore \frac{2}{3}=m_1\times 2and\frac{4}{3}=m_2\times 1$

$\Rightarrow m_1=\frac{1}{3}and{m}_2=\frac{4}{3}$

$\text{Hence, the required lines are}$

$y=\frac{1}{3}xandy=\frac{4}{3}x$

$\Rightarrow x-3y=0and4x-3y=0$