If area of triangle whose vertices are (a, a2),(b, b2),(c, c2) is 1 and area of another triangle whose vertices are (α2, α),(β2, β) and (γ2, γ) is 2 , then the absolute value of |(a+α)2 (b+α)2 (c+α)2 (a+β)2 (b+β)2 (c+β)2 (a+γ)2 (b+γ)2 (c+γ)2|is equal to

Question

If area of triangle whose vertices are (a, a2),(b, b2),(c, c2) is 1 and area of another triangle whose vertices are (α2, α),(β2, β) and (&gamma;2, &gamma;) is 2 , then the absolute value of |(a+α)2 (b+α)2 (c+α)2 (a+β)2 (b+β)2 (c+β)2 (a+&gamma;)2 (b+&gamma;)2 (c+&gamma;)2|is equal to (A) 2 (B) 4 (C) 8 (D) 16

Tardigrade Admin · Accepted Answer

Given determinant =|a 2 2 a 1 b 2 2 b 1 c 2 2 c 1|×|1 α α2 1 β β2 1 &gamma; &gamma;2|=2 × 2 × 1 × 2 × 2=16

Q. If area of triangle whose vertices are (a,a2),(b,b2),(c,c2) is 1 and area of another triangle whose vertices are (α2,α),(β2,β) and (γ2,γ) is 2 , then the absolute value of ∣∣​(a+α)2(a+β)2(a+γ)2​(b+α)2(b+β)2(b+γ)2​(c+α)2(c+β)2(c+γ)2​∣∣​is equal to

2

4

8

16

Given determinant =∣∣​a2b2c2​2a2b2c​111​∣∣​×∣∣​111​αβγ​α2β2γ2​∣∣​=2×2×1×2×2=16

Given determinant $= ∣ ∣ a^{2} b^{2} c^{2} 2 a 2 b 2 c 111 ∣ ∣ \times ∣ ∣ 111 α β γ α^{2} β^{2} γ^{2} ∣ ∣ = 2 \times 2 \times 1 \times 2 \times 2 = 16$