Answer:
an acute triangle
Step-by-step explanation:
Given:
vertex 1 = (7,3)
vertex 2 = (9,0)
vertex 3 = (5,-1)
Now finding the length of each side of the triangle
Using distance formula, to find the length of side between vertex 1 and 2
d=![\sqrt{(x2-x1)^{2}+ (y2-y1)^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B%28x2-x1%29%5E%7B2%7D%2B%20%28y2-y1%29%5E%7B2%7D%20%7D)
Putting values of x1=7 , x2=9, y1=3 and y2=0
d=![\sqrt{(9-7)^{2}+ (0-3)^{2} }\\ =\sqrt{2^{2}+ 3^{2} }\\ =\sqrt{4+9} \\=\sqrt{13}](https://tex.z-dn.net/?f=%5Csqrt%7B%289-7%29%5E%7B2%7D%2B%20%280-3%29%5E%7B2%7D%20%7D%5C%5C%20%3D%5Csqrt%7B2%5E%7B2%7D%2B%203%5E%7B2%7D%20%7D%5C%5C%20%3D%5Csqrt%7B4%2B9%7D%20%5C%5C%3D%5Csqrt%7B13%7D)
Using distance formula, to find the length of side between vertex 1 and 3
Putting values of x1=7 , x2=5, y1=3 and y2=-1
d=![\sqrt{(5-7)^{2}+ (-1-3)^{2} }\\ =\sqrt{2^{2}+ 4^{2} }\\ =\sqrt{4+16} \\=\sqrt{20](https://tex.z-dn.net/?f=%5Csqrt%7B%285-7%29%5E%7B2%7D%2B%20%28-1-3%29%5E%7B2%7D%20%7D%5C%5C%20%3D%5Csqrt%7B2%5E%7B2%7D%2B%204%5E%7B2%7D%20%7D%5C%5C%20%3D%5Csqrt%7B4%2B16%7D%20%5C%5C%3D%5Csqrt%7B20)
Using distance formula, to find the length of side between vertex 2 and 3
Putting values of x1=9 , x2=5, y1=0 and y2=-1
d=![\sqrt{(5-9)^{2}+ (-1-0)^{2} }\\ =\sqrt{4^{2}+ 1^{2} }\\ =\sqrt{16+1} \\=\sqrt{17](https://tex.z-dn.net/?f=%5Csqrt%7B%285-9%29%5E%7B2%7D%2B%20%28-1-0%29%5E%7B2%7D%20%7D%5C%5C%20%3D%5Csqrt%7B4%5E%7B2%7D%2B%201%5E%7B2%7D%20%7D%5C%5C%20%3D%5Csqrt%7B16%2B1%7D%20%5C%5C%3D%5Csqrt%7B17)
Hence the three sides of triangle are:
√13, √20, √17
by Pythagoras theorem
if c^2= a^2 + b^2 then triangle is right triangle
if c^2> a^2 + b^2 then triangle is obtuse triangle
if c^2<a^2 + b^2 then triangle is acute triangle
Now let a=√13 b=√17 and c=√20 then:
a^2 + b^2 = 13+17
= 30
c^2=20
and 20 < 30 which means c^2<a^2 + b^2 then triangle is acute triangle !