Question

Three vertices of a triangle are A (1, 0), B(4, 3) C(3, 2). A point P lies inside the triangle ABC such that [∆PAB] : [∆PBC] : [∆PCA] is 2 : 1 : 3 . (where ∆ABC represents area of triangle ABC )If P is (alpha,beta) then alpha + beta =

Answer 1

Given : Three vertices of a triangle are A(1, 0), B(4, 3) and C(3, 2). point A lies inside the triangle ABC such that ar∆PAB : ar∆PBC : ar∆PCA = 2 : 1 : 3

To find : the value of α + β, if P = (α, β)

solution : ar∆PAB = 1/2 [α(0 - 3) + 1(3 - β) + 4(β - 0)]

= 1/2 [-3α + 3 + 3β ]

ar∆PBC = 1/2 [α(3 - 2) + 4(2 - β) + 3(β - 3)]

= 1/2 [α + 8 - 4β + 3β - 9 ]

= 1/2 [α - β - 1]

ar∆PCA = 1/2 [α(2 - 0) + 3(0 - β) + 1(β - 2)]

= 1/2 [2α - 2β - 2 ]

now, given 1/2 [-3α + 3 + 3β ] : 1/2 [α - β - 1] : 1/2 [2α - 2β - 2] = 2 : 1 : 3

but area of triangle ∆ABC = 1/2 [1(3 - 2) + 4(2 - 0) + 3(0 - 3)]

= 1/2 [1 + 8 - 9 ]

= 0

Therefore it doesn't possible to find out the value of α + β,