Given points are A(a,2a), B(−2,6) and C(3,1). It is also said that the area enclosed by them is 10 square units.

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)

= |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|

Given that area of ∆ABC = 10

∴ 10 = |a(6 – 1) -2 (1 – 2a) + 3(2a - 6)|

∴ 20 = |5a – 2 + 4a + 6a - 18|

∴ 20 = | 15a – 20|

∴ 15a – 20 = ± 20

Taking positive sign,

15a – 20 = 20

a =

Taking negative sign,

15a – 20 = -20

a = 0

Hence, the value of a are 0 and