# 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

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