What is the area of a figure with vertices(1,1), (8,1), and (5,5)?

It will be a triangle
Calculating sides of this triangle
A(1,1)
B(8,1)
C(5,5)
side a-AB=$$\sqrt{(8-1)^{2}+( 1-1)^{2} ) }= \sqrt{ 7^{2} }=7$$
side b-AC=$$\sqrt{(5-1)^{2}+( 5-1)^{2} ) }= \sqrt{ 4^{2}+4^{2} }= \sqrt{36}$$=6 )
side c-BC=$$\sqrt{(5-8)^{2}+( 5-1)^{2} ) }= \sqrt{ (-3)^{2}+4^{2} }= \sqrt{25}$$=5 )
Area of the triangle from Herone formula:
A=$$\sqrt{p(p-a)(p-b)(p-c)}$$
p=$$\frac{1}{2}(a+b+c)$$=$$\frac{1}{2}(7+6+5)$$=9
A=$$\sqrt{9(9-7)(9-6)(9-5)}$$=$$\sqrt{9*2*3*4}$$=$$\sqrt{216}$$
Area is $$\sqrt{216}$$

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