Answer :

Given: equations are as follows:


3x + 2y + 6 = 0 … (1)


2x − 5y + 4 = 0 … (2)


x − 3y − 6 = 0 … (3)


Assuming:


In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.


Concept Used:


Point of intersection of two lines.


Explanation:


Solving (1) and (2):


x = −2, y = 0


Thus, AB and BC intersect at B (−2, 0).


Solving (1) and (3):


x = - 6/11, y = - 24/11


Thus, AB and CA intersect at


Similarly, solving (2) and (3):


x = −42, y = −16


Thus, BC and CA intersect at C (−42, −16).


Let D, E and F be the midpoints the sides BC, CA and AB, respectively. Then,


Then, we have:





Now, the equation of the median AD is



16x – 59y – 120 = 0


The equation of the median BE is



25x – 53y + 50 = 0


And, the equation of median CF is



41 x – 112 y – 70 = 0


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