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.
The line 2x + 3y = 6 intercept the x-axis and the y-axis at A and B, respectively.
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
Clearly, P and Q lie on y = m1x and y = m2x, respectively
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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