The given equations are

y = x …(i)

y = 2x …(ii)

and y – 3x = 4 …(iii)

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

From eq. (i) and (ii), we get x = 0 and y = 0

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

Solving eq. (ii) and (iii), we get

y = 2x …(ii)

and y – 3x = 4 …(iii)

Putting the value of y = 2x in eq. (iii), we get

2x – 3x = 4

⇒ -x = 4

⇒ x = -4

Putting the value of x = -4 in eq. (ii), we get

y = 2(-4)

⇒ y = -8

Thus, BC and AC intersect at (-4, -8)

Now, Solving eq. (iii) and (i), we get

y – 3x = 4 …(iii)

and y = x …(i)

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

x – 3x = 4

⇒ -2x = 4

⇒ x = -2

Putting the value of x = -2 in eq. (i), we get

y = -2

Thus, AC and AB intersect at (-2, -2)

So, vertices of triangle ABC are: (0, 0), (-4, -8) and (-2, -2)

= 4 sq. units