Question

Find the area of a triangle whose vertices are (6,3)(-3,5)(4,-2)

Answer 1

Answer:

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Answer 2

$\huge\bold{\underline{Question:}}$

Find the area of a triangle whose vertices are (6,3)(-3,5)(4,-2)

$\huge\bold{\underline{Answer:}}$

Given:

Let A(6, 3), B(-3, 5) and C(4,-2) be the given points.

To find:

Find the area of a triangle

Solution:

Vertices of the given triangle are A(6, 3), B(-3, 5) and C(4,-2)

We know that,

Area of a triangle = $\boxed{\bf{\pink{\dfrac{1}{2}×[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]}}}$

Here,

• x₁ = 6

• y₁ = 3

• x₂ = -3

• y₂ = 5

• x₃ = 4

• y₃ = -2

Area of the triangle:

$\bf{\dfrac{1}{2}×[3(-3-4)+5(4-6)+(-2)(6+3)]}$

$\bf{\dfrac{1}{2}×[-21-10-18]}$

$\bf{\dfrac{1}{2}×49}$

$\boxed{\bf{\pink{\dfrac{49}{2}=24.5\:sq.units}}}$

Therefore,

Therefore,Area of the triangle is 24.5 sq.units.

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