The given line is
3x + 4y = 12
⇒ \(\frac{x}{4}\) + \(\frac{y}{3}\) = 1
Line (1) Cuts x-axis and y-axis at A and B respectively. Then A is (4, 0) and B is (0, 3).
Let the line AB be trisected at P and Q, then
AP: PB = 1: 2 and AQ: QB = 2: 1
∴ = \(\big(\frac{1.0 + 2.4}{1 + 2}, \frac{1.3 + 2.0}{1 + 2}\big)\) = (\(\frac{8}{3}\), 1)
And Q = \(\big(\frac{2.0 + 1.4}{1 + 2}, \frac{2.3 + 1.0}{1 + 2}\big)\) = (\(\frac{4}{3}\), 2)
Now, equation of line OP passing through (0, 0) and (\(\frac{8}{3}\), 1) is
y − 0 = \(\frac{1-0}{\frac{8}{3}-0}\) (x − 0) ⇒ y = \(\frac{3}{8}\) x ⇒ 3x − 8y = 0
∴ The Equation of the straight line is
3x – 8y = 0 or 3x – 2y = 0
