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To find: Equations of the straight lines which pass through the origin and trisect the portion of the line which is intercepted between the axes.

Assuming:

The line 2x + 3y = 6 intercept the x-axis and the y-axis at A and B, respectively.

Explanation:

At x = 0 we have,

3y + 0 = 6

3y = 6

y = 2

At y = 0 we have,

2x + 0 = 6

x = 3

A = (3, 0) and B = (0, 2)

Let y = m1x and y = m2x pass through origin trisecting the line 2x + 3y = 6 at P and Q.

AP = PQ = QB

Let us find the coordinates of P and Q using the section formula

P Q Clearly, P and Q lie on y = m1x and y = m2x, respectively  and Hence, the required lines are

y nd y x – 3y = 0 and 4x – 3y = 0

Hence, the equation of line is x – 3y = 0 and 4x – 3y = 0

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