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Determine, algebraically, the vertices of the triangle formed by the lines 

3x – y = 3  

2x – 3y = 2  

x + 2y = 8

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3x – y = 2 …(i)

2x -3y = 2 …(ii)

x + 2y = 8 …(iii)

Let the equations of the line (i), (ii) and (iii) represent the side of a ∆ABC.

On solving (i) and (ii),

We get,

[First, multiply Eq. (i) by 3 in Eq. (i) and then subtract]

(9x-3y)-(2x-3y) = 9-2

7x = 7

x = 1

Substituting x=1 in Eq. (i), we get

3×1-y = 3

y = 0

So, the coordinate of point B is (1, 0)

On solving lines (ii) and (iii),

We get,

[First, multiply Eq. (iii) by 2 and then subtract]

(2x + 4y)-(2x-3y) = 16-2

7y = 14

y = 2

Substituting y=2 in Eq. (iii), we get

x + 2 × 2 = 8

x + 4 = 8

x = 4

Hence, the coordinate of point C is (4, 2).

On solving lines (iii) and (i),

We get,

[First, multiply in Eq. (i) by 2 and then add]

(6x-2y) + (x + 2y) = 6 + 8

7x = 14

x = 2

Substituting x=2 in Eq. (i), we get

3 × 2 – y = 3

y = 3

So, the coordinate of point A is (2, 3).

Hence, the vertices of the ∆ABC formed by the given lines are as follows,

A (2, 3), B (1, 0) and C (4, 2).

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