Question

let be three points.then find area of the following triangles ​

Answer 1

Hlo mate here is ur ans....

Let (x₁, y₁), (x₂, y₂) and (x₃, y₃) be the co-ordinates of the vertices A, B, C respectively of the triangle ABC. We are to find the area of the triangle ABC...

Draw AL, BM and CN perpendiculars from A, B and C respectively on the x-axis.

Then, we have, OL = x₁, OM = x₂, ON = x₃ and AL = y₁, BM = y₂, CN = y₃.

Then, we have, OL = x₁, OM = x₂, ON = x₃ and AL = y₁, BM = y₂, CN = y₃.Therefore, LM = OM - OL = x₂ - x₁;NM = OM - ON = x₂ - x₃;and LN = ON - OL = x₃ - x₁.Since the area of a trapezium =1/2× the sum of the parallel sides × the perpendicular distance between them,

× the sum of the parallel sides × the perpendicular distance between them, Hence, the area of the triangle ABC = ∆ABC

= area of the trapezium ALNC + area of the trapezium CNMB - area of the trapezium ALMB

= area of the trapezium ALNC + area of the trapezium CNMB - area of the trapezium ALMB = 1/2 ∙ (AL + NC) . LN + 1/2 ∙ (CN + BM) ∙ NM - 1/2 ∙ (AL + BM).LM

2 ∙ (AL + BM).LM= 1/2 ∙ (y₁ + y₃) (x₃ - x₁) + 1/2 ∙ (y₃ + y₂) (x₂ - x₃) - 12 ∙ (y₁ + y₂) (x₂ - x₁)

2 ∙ (y₃ + y₂) (x₂ - x₃) - 12 ∙ (y₁ + y₂) (x₂ - x₁)= 1/2 ∙ [x₁ y₂ - y₁ x₂ + x₂ y₃ - y₂ x₃ + x₃ y₁ - y₃ x₁]

2 ∙ [x₁ y₂ - y₁ x₂ + x₂ y₃ - y₂ x₃ + x₃ y₁ - y₃ x₁] = 12[x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)] sq. units..

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