The given equations are

3x + 2y + 6 = 0 …(i)

2x – 5y + 4 = 0 …(ii)

and x – 3y – 6 = 0 …(iii)

Let eq. (i), (ii) and (iii) represents the sides AB, BC and AC respectively of ΔABC

Firstly, we solve the equation (i) and (ii)

3x + 2y + 6 = 0 …(i)

2x – 5y + 4 = 0 …(ii)

Multiplying the eq. (i) by 2 and (ii) by 3, we get

6x + 4y + 12 = 0 …A

6x – 15y + 12 = 0 …B

Subtracting eq. (B) from (A), we get

6x + 4y + 12 – 6x + 15y – 12 = 0

⇒ 19y = 0

⇒ y = 0

Putting the value of y = 0 in eq. (i), we get

3x + 2(0) + 6 = 0

⇒ 3x + 6 = 0

⇒ 3x = -6

⇒ x = -2

Thus, AB and BC intersect at (-2, 0)

Now, we solve eq. (ii) and (iii)

2x – 5y + 4 = 0 …(ii)

and x – 3y – 6 = 0 …(iii)

Multiplying the eq. (iii) by 2, we get

2x – 6y – 12 = 0 …(iv)

Subtracting eq. (iv) from (ii), we get

2x – 5y + 4 – 2x + 6y + 12 = 0

⇒ y + 16 = 0

⇒ y = -16

Putting the value of y = -16 in eq. (ii), we get

2x – 5(-16) + 4 = 0

⇒ 2x + 80 + 4 = 0

⇒ 2x + 84 = 0

⇒ 2x = -84

⇒ x = -42

Thus, BC and AC intersect at (-42, -16)

Now, we solve eq. (iii) and (i)

x – 3y – 6 = 0 …(iii)

3x + 2y + 6 = 0 …(i)

Multiplying the eq. (iii) by 3, we get

3x – 9y – 18 = 0 …(v)

Subtracting eq. (v) from (i), we get

3x + 2y + 6 – 3x + 9y + 18 = 0

⇒ 11y + 24 = 0

⇒ 11y = -24

Putting the value of y in eq. (iii), we get

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

Then the coordinates of D, E and F are

= (-22, -8)

Now, we have to find the equations of Medians AD, BE and CF

The equation of median AD is

⇒ 16x – 59y = 120

The equation of the median BE is

⇒ 58y = 25x + 50

⇒ 25x – 58y + 50 = 0

The equation of median AD is

⇒ 112y + 1792 = 41x + 1722

⇒ 41x – 112y + 1722 – 1792 = 0

⇒ 41x – 112y – 70 = 0