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:
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.