Sum
Find the area of triangle whose vertices are
M(0, 5), N(−2, 3), T(1, −4)
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Solution
Here, M(x1, y1) ≡ M(0, 5), N(x2, y2) ≡ N(−2, 3), T(x3, y3) ≡ T(1, −4)
Area of a triangle = `1/2|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`
∴ A(ΔMNT) = `1/2|(0, 5, 1),(-2, 3, 1),(1, -4, 1)|`
= `1/2[0 - 5 (-2 - 1) + 1(8 - 3)]`
= `1/2[-5 (-3) + 1(5)]`
= `1/2(15 + 5)`
= `1/2(20)`
= 10 sq. units
Concept: Application of Determinants - Area of Triangle and Collinearity of Three Points
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