Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9)

Solution:

Given, the vertices of the triangle are (-8, 4) (-6, 6) and (-3, 9)

We have to find the area of the triangle.

The area of a triangle with vertices A (x₁ , y₁) , B (x₂ , y₂) and C (x₃ , y₃) is

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

Here, (x₁ , y₁) = (-8, 4) (x₂ , y₂) = (-6, 6) and (x₃ , y₃) = (-3, 9)

Area of the triangle = 1/2[-8(6 - 9) + -6(9 - 4) + -3(4 - 6)]

= 1/2[-8(-3) - 6(5) - 3(-2)]

= 1/2[24 - 30 + 6]

= 1/2[-6 + 6]

= 0

Therefore, the area of the triangle = 0 square units.

✦ Try This: The area of a triangle with vertices A(3,0), B(0, 7) and C(0, 4) is

Given, the vertices are A(3, 0) B(0, 7) and C(0, 4)

The area of a triangle with vertices A (x₁ , y₁) , B (x₂ , y₂) and C (x₃ , y₃) is

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

Here, (x₁ , y₁) = (3, 0) (x₂ , y₂) = (0, 7) and (x₃ , y₃) = (0, 4)

Area of the triangle = 1/2[3(7 - 4) + 0(4 - 0) + 0(0 - 7)]

= 1/2[3(3) + 0 + 0]

= 1/2[9 + 0]

= 9/2

= 4.5 square units.

Therefore, the area of the triangle is 4.5 square units.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.3 Problem 9

Summary:

The area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9) is zero

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