Let the three vertices of triangle ABC be:

A (2, 5), B (-2, 2) and C (4, 2)

Now, when we plot these points in the graph paper then we see that,

Point A and C lie in the quadrant I and point B lie in the II quadrant

Let the line BC intersect y-axis at point D

∴ BC = (BD + DC)

= (2 + 4) units

= 6 units

Now, we have to draw AM perpendicular to x-axis and intersect BC at L

∴ Ordinate of point L = Ordinate of point B - Ordinate of point C

AL = AM – LM

= 5 – 2

= 3 units

Hence, Area of triangle ABC =

=

=

= 9 square units

∴ Area of triangle ABC = 9 square units