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

3x-y = 2

2x-3y = 2

and x + 2y = 8

Given equation of lines are

3x - y = 2 …(i)

2x -3y = 2 …(ii)

and x + 2y = 8 …(iii)

Let lines (i), (ii) and (iii) represent the side of a ∆ABC i.e., AB, BC and CA respectively.

On solving lines (i) and (ii), we will get the intersecting point B.

On multiplying Eq. (i) by 3 in Eq. (i) and then subtracting, we get

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

7x = 7

x = 1

On putting the value of x in Eq. (i), we get

3×1-y = 3
y = 0

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

On solving lines (ii) and (iii), we will get the intersecting point C.

On multiplying Eq. (iii) by 2 and then subtracting, we get

(2x + 4y)-(2x-3y) = 16-2
7y = 14
y = 2

On putting the value of y in Eq. (iii), we get

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

On solving lines (iii) and (i), we will get the intersecting point A.

On multiplying in Eq. (i) by 2 and then adding Eq. (iii), we get

(6x-2y) + (x + 2y) = 6 + 8
7x = 14
x = 2

On putting the value of x in Eq. (i), we get

3×2 - y = 3
y = 3

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

Hence, the vertices of the ∆ABC formed by the given lines are A (2, 3), B(1, 0) and C (4, 2).