Question

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

3x – y = 2

2x – 3y = 2

x + 2y = 8

Answer 1

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 equation (i) by 3 in equation (i) and then subtract

(9x – 3y) – (2x – 3y) = 9 – 2

7x = 7

x = 1

Substituting x = 1 in equation (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 equation (iii) by 2 and then subtract

(2x + 4y) – (2x – 3y) = 16 – 2

7y = 14

y = 2

Substituting y = 2 in equation (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 equation (i) by 2 and then add

(6x – 2y) + (x + 2y) = 6 + 8

7x = 14

x = 2

Substituting x = 2 in equation (i), we get

3 × 2 – y = 3

y = 6 – 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).