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).