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

$\text{The required straight line passes through (0, 0) and trisects the part of the line}3x+y=12\text{that lies between the axes of coordinates}\text{The line 3x + y = 12 has A(4, 0) and B(0, 12) as x and y intercepts.}\text{Let P and Q be the points of trisection of AB.}\text{Since P divides AB in the ration 1 : 2, coordinates of P are :}P=\frac{1\left(0\right)+2\left(4\right)}{1+2}=\frac{1\left(12\right)+2\left(0\right)}{1+2}=(\frac{8}{3},4)\text{Since Q divides BA in the ratio 1 : 2, coordinates of Q are :}Q=\frac{2\left(0\right)+1\left(4\right)}{1+2}=\frac{1\left(0\right)+2\left(12\right)}{1+2}=(\frac{4}{3},8)\text{Equation of line through (0, 0) and P}(\frac{8}{3},4)is:y-0=\frac{4-0}{\frac{8}{3}-0}(x-0)\frac{4\times 3}{8}xy-0=\frac{12}{8}x2y=3x\text{Equation of line through (0, 0) and}Q(\frac{4}{3},8)is:y-0=\frac{8-0}{\frac{4}{3}-0}(x-0)=6x=\frac{8\times 3}{4}\left(x\right)y=6x$