The area of the triangle whose vertices are (2,-3) (3,2) and (-2,5) is​

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1. Given:-

   • Vertices of the triangles:-
   • (2, -3), (3, 2) and (-2, 5)

   To Find:-

   • The area of the triangle.

   Solution:-

   Firstly we have the given points as:-

   • (2, -3)
   • (3, 2)
   • (-2, 5)

   From these points we get the following:-

   • x₁ = 2
   • x₂ = 3
   • x₃ = -2
   • y₁ = -3
   • y₂ = 2
   • y₃ = 5

   We already know:–

   • Area of triangle = 1/2[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

   Putting all the values in the formula:-

   Area = 1/2[2(2 – 5) + 3{5 – (-3)} + {-2(-3 – 2)}]

   = Area = 1/2[2 × (-3) + 3(5 + 3) + {(-2) × (-5)}]

   = Area = 1/2[-6 + 3 × 8 + 10]

   = Area = 1/2[-6 + 24 + 10]

   = Area = 1/2[-6 + 34]

   = Area = 1/2[28]

   = Area = 14 sq.units

   ∴ Area of the triangle is 14 sq.units

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   Important Points:-

   We need to remember the meaning of x₁, x₂ …. y₁, y₂.

   So:-

   • x₁ = abscissa of the 1st point
   • x₂ = abscissa of the 2nd point
   • x₃ = abscissa of the 3rd point
   • y₁ = ordinate of the 1st point
   • y₂ = ordinate of the 2nd point
   • y₃ = ordinate of the 3rd point

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