Respuesta :

4143689097

Answer:

[tex]Area = 51\:unit^2[/tex]

Step-by-step explanation:

[tex]Area = \sqrt{p(p-a)(p-b)(p-c)}\quad and \quad p=\frac{a+b+c}{2}[/tex]

Find the distance between three points.

[tex]\overline{AB}=\sqrt{\left(-4-\left(-5\right)\right)^2+\left(-5-7\right)^2}=12\\\\\overline{BC}=\sqrt{\left(4-\left(-4\right)\right)^2+\left(5-\left(-5\right)\right)^2}=12.8\\\\\overline{AC}=\sqrt{\left(4-\left(-5\right)\right)^2+\left(5-7\right)^2}=9.2[/tex]

[tex]p=\frac{12+12.5+9.2}{2}=16.9[/tex]

[tex]Area = \sqrt{16.9\left(16.9-12\right)\left(16.9-12.8\right)\left(16.9-9.2\right)}\\\\=51\:unit^2[/tex]

Best Regards!

TheAnimeGirl

Answer:

53 sq.units

solution,

A(-5,7)--->(X1,y1)

B(-4,-5)-->(x2,y2)

C(4,5)--->(X3,y3)

Now,

Area of triangle:

[tex] \frac{1}{2} (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) \\ = \frac{1}{2} ( - 5( - 5 - 5) + ( - 4)(5 - 7) + 4(7 - ( - 5) \\ = \frac{1}{2} ( - 5 \times ( - 10) + ( - 4) \times ( - 2) + 4 \times 12) \\ = \frac{1}{2} (50 + 8 + 48) \\ = \frac{1}{2} \times 106 \\ = \frac{106}{2} \\ = 53[/tex]

hope this helps...

Good luck on your assignment..

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