Find the eq. of the straight lines which go through the origin and trisect the portion of the straight line 3x+y=12 which is intetercepted between the axes of the coordinates.

**Given** line is:

3x + y = 12

Its point of intersection on the graph can be find out as:

x | 0 | 4 |

y | 12 | 0 |

So, the point through which the given line passes are let A(4, 0) and B(0, 12).

Let P and Q be the point of intersection trisecting AB.

Here, P divides AB in the ratio 1 : 2.

So, the co-ordinates of P

Given, point O is origin whose co-ordinates must be (0, 0).

Therefore, equation of line OP is:

Similarly, the point Q will divide AB in the ratio 2 : 1, so its coordinates are

co-ordinates of Q

Again, the equation of OQ is:

Hence, equation (1) and (2) are the required equation of the line.

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