Answer:
[tex](BC)^2=(AC)^2+(AB)^2[/tex]. It means AC is perpendicular to AB.
Step-by-step explanation:
Vertices of given triangle are A(−2, −2), B(5, −2), and C(−2,22).
Distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using distance formula we get
[tex]AB=\sqrt{(5-(-2))^2+(-2-(-2))^2}=\sqrt{7^2}=7[/tex]
[tex]BC=\sqrt{(-2-5)^2+(22-(-2))^2}=\sqrt{7^2+24^2}=\sqrt{625}=25[/tex]
[tex]AC=\sqrt{(-2-(-2))^2+(22-(-2))^2}=\sqrt{24^2}=24[/tex]
According to Pythagoras theorem
[tex]hypotenuse^2=perpendicular^2+base^2[/tex]
[tex](BC)^2=(AC)^2+(AB)^2[/tex]
Perpendicular = AC
Base = AB
It means AC is perpendicular to AB.
Hence proved.
