Question

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Answer 1

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula:
`d= \sqrt{(x_(2)-x_(1))^2+(y_(2)-y_(1))^2}`

Distance from point A to point B:

`d_(AB)= √([6-(-9)]^2+(10-5)^2)`

`d_(AB)= √((6+9)^2+(10-5)^2)`

`d_(AB)= √((15)^2+(5)^2)`

`d_(AB)= √(225+25)`

`d_(AB)= √(250)`

`d_(AB)=15.81`

Distance from point A to point C:

`d_(AC)= √([2-(-9)]^2+(-10-5)^2)`

`d_(AC)= √((2+9)^2+(-10-5)^2)`

`d_(AC)= √(11^2+(-15)^2)`

`d_(AC)= √(121+225)`

`d_(AC)= √(346)`

`d_(AC)= 18.60`

Distance from point B from point C

`d_(BC)= √((2-6)^2+(-10-10)^2)`

`d_(BC)= √((-4)^2+(-20)^2)`

`d_(BC)= √(16+400)`

`d_(BC)= √(416)`

`d_(BC)=20.40`

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:

`s= (AB+AC+BC)/(2)`

`s= (15.81+18.60+20.40)/(2)`

`s= (54.81)/(2)`

`s=27.41`

Finally, to find the area of our triangle, we are going to use Heron's formula:

`A= √(s(s-AB)(s-AC)(s-BC))`

`A=√(27.41(27.41-15.81)(27.41-18.60)(27.41-20.40))`

`A= √(27.41(11.6)(8.81)(7.01))`

`A=140.13`

We can conclude that the perimeter of our triangle is 140.13 square units.