Question

Find the area of the triangle whose vertices are A(4,4)'B(3,-16)and C(3,-2)

Answer 1

soln: The area of the triangle with vertices (a₁, b₁), (a₂, b₂) & (a₃, c₃) is given by

Δ = ±1/2 Ι$\left[\begin{array}{ccc}a₁&b₁&1\\a₂&b₂&1\\a₃&b₃&1\end{array}\right]$Ι

Here, (a₁, b₁)  ≡  (4, 4), (a₂, b₂)  ≡  (3, -16) & (a₃, b₃)  ≡  (3, -2)

=> Δ = ± 1/2Ι$\left[\begin{array}{ccc}4&4&1\\3&-16&1\\3&-2&1\end{array}\right]$Ι

Note: Modulus of Matrix is nothing but determinant(i,e |[A]| = det(A))

=> Δ = ±1/2{4(-16 + 2) -4(3 - 3) +1(-6 + 48)}

=> Δ = ±1/2{-56 - 0 + 42} = ± 7

Remarks:
Since area is a positive quantity, we always take the absolute value of the determinant

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Method 2: 

Area of ΔABC = | 1/2[(x₂ - x₁)(y₃ - y₁) - (x₃ - x₁)(y₂ - y₁)] |

Area of ΔABC = | 1/2[(3 - 4)(-2 - 4) - (3 - 4)(-16 - 4)] | = | 1/2([ 6 - 20] | = | -7 | = 7

Hope it helps you!