The area of the triangle formed by (a, 2a), (–2, 6) and (3, 1) is 5 units.  Then a =

A(a,2a),B(−2,6),C(3,1) Area of ΔABC=12|a+22a−6a−32a−1|=5⇒|(a+2)(2a−1)−(a−3)(2a−6)|=10⇒|(3a−2)+12a−18|=10⇒|15a−20|=10|3a−4|=2     a=2or 23