Answer:
Yes, It is a right triangle for the given vertices.
Step-by-step explanation:
Given:
Q(7, –10),
R(–3, 0),
S(9, –8)
To Find:
determine whether is a rig ht triangle for the given vertices = ?
Solution:
QR=[tex]\sqrt{(-3-7)^2+(0-(-10))^2}[/tex]
QR=[tex]\sqrt{(-10)^2+(10)^2}[/tex]
QR=[tex]\sqrt{100+100}[/tex]
QR=[tex]\sqrt{200}[/tex]--------------------------(1)
QR=14.142136
RS=[tex]\sqrt{(9-(-3))^2+(-8-0)^2}[/tex]
RS=[tex]\sqrt{(12)^2+(-8)^2}[/tex]
RS=[tex]\sqrt{144+64}[/tex]
RS=[tex]\sqrt{208}[/tex]---------------------------(2)
RS=14.422205
QS=[tex]\sqrt{(7-9)^2+(-10-(-8))^2}[/tex]
QS=[tex]\sqrt{(-2)^2+(-2)^2}[/tex]
QS=[tex]\sqrt{4+4}[/tex]
QS=[tex]\sqrt{8}[/tex]-------------------------------------(3)
QS=2.828427
According to Pythagorean Theorem,
[tex]RS^2 = QR^ +QS^2[/tex]
Substituting the values,
[tex](\sqrt{208})^2 = (\sqrt{200})^2 +(\sqrt{8})^2[/tex]
[tex]208 = 200 +8[/tex]
208 = 208
Pythagorean theorem is satisfied. Hence it is a right triangle.
