Question

Two straight lines drawn through the origin, trisect the portion of the line 4x+3y=12 intercepted between the axes. Find the equation of the lines

Answer 1

Answer:

$3y-2x=0$ and $3y-8x=0$

Step-by-step explanation:

Given equation of the line

$4x+3y=12$

at x = 0

$0+3y=12$

$\implies y=4$

at y = 0

$4x+0=12$

$\implies x=3$

Thus the line intersects the x and y axis at points (3, 0) and (0, 4) respectively

Let the points $(x_1,y_1)$ and $(x_2,y_2)$ are the trisection points

These points will divide the line internally in the ratio 1:2

Thus

$x_1=\frac{1\times0+2\times3}{1+2}$

$\implies x_1=\frac{6}{3}$

$\implies x_1=2$

$y_1=\frac{1\times4+2\times0}{1+2}$

$\implies y_1=\frac{4}{3}$

Similarly

$x_2=\frac{1\times3+2\times0}{1+2}$

$\implies x_1=\frac{3}{3}$

$\implies x_1=1$

$y_2=\frac{1\times0+2\times4}{1+2}$

$\implies y_1=\frac{8}{3}$

Thus the coordinates of the trisection points are $(2,4/3)$ and $(1,8/3)$

Equation of line passing through the points $(0,0)$ and $(2,4/3)$

$y-0=\frac{4/3-0}{2-0}(x-0)$

$\implies y=\frac{2}{3}x$

$\implies 3y-2x=0$

Equation of line passing through the points $(0,0)$ and $(1,8/3)$

$y-0=\frac{8/3-0}{1-0}(x-0)$

$\implies y=\frac{8}{3}x$

$\implies 3y-8x=0$

Therefore the equations of the lines are

$3y-2x=0$ and $3y-8x=0$

Hope this helps.