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What is the area of a triangle with vertices at (−2, 1) , (2, 1) , and ​ (3, 4) ​ ?

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What is the area of a triangle with vertices at 2 1 2 1 and 3 4 Enter your answer in the box class=

Respuesta :

Given

A triangle.

with vertices at (−2, 1) , (2, 1) , and ​ (3, 4) ​

Find out  the area of a triangle.

To proof

Formula

[tex]Area\ of\ triangle = \frac{1}{2}[ x_{1} (y_{2} -y_{3} ) + x_{2} (y_{3} - y_{1})+x_{3}(y_{1}-y_{2})[/tex]

As given the vertices at (−2, 1) , (2, 1) , and ​ (3, 4)

put in the above equation

we get

[tex]= \frac{1}{2} [-2(1-4)+ 2 (4-1) + 3 ( 1-1) ][/tex]

solving

[tex]= \frac{1}{2} [6 + 6][/tex]

thus

[tex]=\frac{1}{2} [12][/tex]

area of the triangle is 6 units².

Hence proved

The area of a triangle with vertices at (−2, 1) , (2, 1) , and (3, 4) is; Area = 6 units²

The formula for the area of a triangle when given the 3 vertices is;

Area = ½[Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By)]

In this question, the vertices coordinates are; A(−2, 1), B(2, 1), and C(3, 4)

Thus;

Ax = -2

Bx = 2

Cx = 3

Ay = 1

By = 1

Cy = 4

Plugging in the relevant values into the area equation gives;

Area = ½[-2(1 - 4) + 2(4 - 1) + 3(1 - 1)]

Area = ½(6 + 6 + 0)

Area = ½ × 12

Area = 6 units²

Read more at; https://brainly.com/question/19044159

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