Answer :

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

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