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What is the area of a triangle whose vertices are D(3, 3) , E(3, −1) , and F(−2, −5) ?

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__units²

Respuesta :

Answer        

Find out the Area of a triangle .

To proof  

Formula

Area of Triangle

[tex]= \frac{1}{2}\times( {x_1(y_{2}-y_{3}) +x_{2}(y_{3} - y_{1})+x_{3}(y_{1}-y_{2})})[/tex]

Now vertices are D(3, 3) , E(3, −1) , and F(−2, −5) .

[tex]= \frac{1}{2} (3\times(-1 +5) + 3\times(-5-3)-2\times(3+1))[/tex]

Solving the above

[tex]= \frac{1}{2}(3\times4+3\times-8 -2\times4)[/tex]

[tex]=\frac{1}{2} (12-24-8)\\ =\frac{1}{2} (-32+12)\\=\frac{1}{2} (-20)[/tex]

= -10 units²

(Neglected the negative sign as area cannot be negative.)

= 10 units ²

Area of  a triangle is 10 units ²

Hence proved

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