Answer:
the area of ∆ABC = 9
Step-by-step explanation:
[tex]CA=\sqrt{\left( -1-2\right)^{2} +\left( 2-5\right)^{2} } = \sqrt{18} =3\sqrt{2}[/tex]
[tex]CB=\sqrt{\left( 5-2\right)^{2} +\left( 2-5\right)^{2} } = \sqrt{18} =3\sqrt{2}[/tex]
Since CA = CB then A(-1,2), B (5,2) and C (2,5) are the vertices of an isosceles triangle.
Area calculation:
Let The point H be the midpoint of side AB
[tex]x_{H}=\frac{-1+5}{2} =2\\y_{H}=\frac{2+2}{2} =2[/tex]
Then
H(2 , 2)
therefore,
[tex]CH=\sqrt{\left( 2-2{}\right)^{2} +\left( 2-5\right)^{2} } =3[/tex]
Finally,
[tex]\text{the area of triangle ABC} =\frac{\text{CH} \times \text{AB} }{2} =\frac{3\times 6}{2} =9[/tex]
