Find the equations to the straight lines which go through the origin and trisect the portion of the straight line 3 x + y = 12 which is intercepted between the axes of coordinates.
To find:
The equation of the required line.
Assuming:
The line 3x + y = 12 intersect the x-axis and the y-axis at A and B, respectively.
y =m1x and m2x be the lines passing through the origin and trisect the line 3x + y =12 at P and Q.
Explanation:
At x = 0
0 + y =12
⇒ y =12
At y = 0
3x + 0 =12
⇒ x = 4
A (4, 0) and B (0,12)
y =m1x and m2x be the lines passing through the origin and trisect the line 3x + y =12 at P and Q.
AP = PQ = QB
Let us find the coordinates of P and Q.
P 
Q
Clearly, P and Q lie on y = m1x and y = m2x, respectively.
4 
⇒ m1
and m2 = 6
Hence, the required lines are y 
⇒ 2y = 3x and y = 6x
Hence, the equation of line is 2y = 3x and y = 6x