Question

Find the area of triangle ABC whose vertices are
A (-5, 7), B (-4, -5) and C (4, 5).

Answer 1

`Area = 51\:unit^2`

Explanation:

`Area = √(p(p-a)(p-b)(p-c))\quad and \quad p=(a+b+c)/(2)`

Find the distance between three points.

`\overline{AB}=√(\left(-4-\left(-5\right)\right)^2+\left(-5-7\right)^2)=12\\\\\overline{BC}=√(\left(4-\left(-4\right)\right)^2+\left(5-\left(-5\right)\right)^2)=12.8\\\\\overline{AC}=√(\left(4-\left(-5\right)\right)^2+\left(5-7\right)^2)=9.2`

`p=(12+12.5+9.2)/(2)=16.9`

`Area = √(16.9\left(16.9-12\right)\left(16.9-12.8\right)\left(16.9-9.2\right))\\\\=51\:unit^2`

Best Regards!

Answer 2

53 sq.units

solution,

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

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

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

Now,

Area of triangle:

`(1)/(2) (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) \\ = (1)/(2) ( - 5( - 5 - 5) + ( - 4)(5 - 7) + 4(7 - ( - 5) \\ = (1)/(2) ( - 5 * ( - 10) + ( - 4) * ( - 2) + 4 * 12) \\ = (1)/(2) (50 + 8 + 48) \\ = (1)/(2) * 106 \\ = (106)/(2) \\ = 53`

hope this helps...

Good luck on your assignment..