Q. 174.0( 6 Votes )

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Answer :

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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