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=[tex]\sqrt{(x2-x1)^{2}+ (y2-y1)^{2} }[/tex]
Putting values of x1=7 , x2=9, y1=3 and y2=0
d=[tex]\sqrt{(9-7)^{2}+ (0-3)^{2} }\\ =\sqrt{2^{2}+ 3^{2} }\\ =\sqrt{4+9} \\=\sqrt{13}[/tex]
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=[tex]\sqrt{(5-7)^{2}+ (-1-3)^{2} }\\ =\sqrt{2^{2}+ 4^{2} }\\ =\sqrt{4+16} \\=\sqrt{20[/tex]
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=[tex]\sqrt{(5-9)^{2}+ (-1-0)^{2} }\\ =\sqrt{4^{2}+ 1^{2} }\\ =\sqrt{16+1} \\=\sqrt{17[/tex]
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 !
