Answer:
20
Step-by-step explanation:
firstly let's find out the side lengths of the triangle using formula
[tex]length \: = \: \sqrt{(y₂-y1) + ( y₂-y1)}[/tex]
length of AB
[tex] \sqrt{( - 6 - 2) {}^{2} + ( - 4 - 1) {}^{2} } = \sqrt{89} [/tex]
length of AC
[tex] \sqrt{( - 6 - 2) {}^{2} + (1 - 1) {}^{2} } = 8[/tex]
length of BC
[tex] \sqrt{( - 6 - ( - 6)) {}^{2} + (1 - ( - 4)) {}^{2} } = 5[/tex]
Now we have all three side lengths:
a =√89 b = 8 and c = 5
using this formula to find the semi perimeter where a, b and c are the side lengths, input them in
[tex]s \: = \: \frac{a + b + c}{2} [/tex]
[tex]s = \frac{ \sqrt{89} + 8 + 5 }{2} = \frac{13 + \sqrt{89} } {2} [/tex]
Use heron's formula
[tex]area = \sqrt{s(s - a)(s - b)(s - c)} [/tex]
[tex]area = \sqrt{ \frac{13 + \sqrt{89} }{2}(\frac{13 + \sqrt{89} }{2} - \sqrt{89})( \frac{13 + \sqrt{89} }{2} - 8)(\frac{13 + \sqrt{89} }{2} - 5 } = 20[/tex]
area = 20
