Let P and Q be the points of trisection of the line segment joining the points A (4, -1) and B (-2, -3), where P is nearest to A. If the coordinates of P and Q are (α, β) and (γ, δ) respectively, then the value of (α + 2β - 3γ - δ) is:

Given:

A (4, -1)

B (-2, -3)

P and Q are points of trisection of AB, P is nearest to A.

P (α, β)

Q (γ, δ)

Formula used:

Section formula: If a point divides a line segment joining (x₁, y₁) and (x₂, y₂) in the ratio m:n, then the coordinates are:

x = (mx₂ + nx₁) / (m + n)

y = (my₂ + ny₁) / (m + n)

Calculation:

The coordinates of P:

P divides AB in the ratio 1:2.

α = (1 × (-2) + 2 × 4) / (1 + 2)

⇒ (-2 + 8) / 3 = 6 / 3 = 2

β = (1 × (-3) + 2 × (-1)) / (1 + 2)

⇒ (-3 - 2) / 3 = -5 / 3

P (2, -5/3)

The coordinates of Q:

Q divides AB in the ratio 2:1.

γ = (2 × (-2) + 1 × 4) / (2 + 1)

⇒ (-4 + 4) / 3 = 0 / 3 = 0

δ = (2 × (-3) + 1 × (-1)) / (2 + 1)

⇒ (-6 - 1) / 3 = -7 / 3

Q (0, -7/3)

Now for, α + 2β - 3γ - δ:

α + 2β - 3γ - δ = 2 + 2(-5/3) - 3(0) - (-7/3)

⇒ 2 - 10/3 + 7/3 

⇒ 2 - 3/3 = 1

∴ The value of (α + 2β - 3γ - δ) is 1.