Answer:
The answer is "True".
Explanation:
Given points:
[tex]A= (-6,-8)\\B=(-16, 12)\\C=(-26, -18)\\[/tex]
Condition of the right-angle triangle is: [tex]\bold{AC^2= AB^2+BC^2}[/tex]
Formula to find length:
[tex]Length= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]AC = (-6,-8)(-26,-18)\\\\x_1= -6 \ \ y_1=-8 \ \ \ and \ \ \ x_2=-26 \ \ y_2=-18\\[/tex]
[tex]\Rightarrow AC= \sqrt{(-26-(-6))^2+(-18-(-8))^2}\\\\\Rightarrow AC= \sqrt{(-26+6))^2+(-18+8))^2}\\\\\Rightarrow AC= \sqrt{(-20))^2+(-10))^2}\\\\\Rightarrow AC= \sqrt{400+100}\\\\\Rightarrow AC= \sqrt{500}\\\\[/tex]
[tex]AB = (-6,-8)(-16,12)\\\\x_1= -6 \ \ y_1=-8 \ \ \ and \ \ \ x_2=-16 \ \ y_2=12\\[/tex]
[tex]\Rightarrow AB= \sqrt{(-16-(-6))^2+(12-(-8))^2}\\\\\Rightarrow AB= \sqrt{(-16+6))^2+(12+8))^2}\\\\\Rightarrow AB= \sqrt{(-10))^2+(20))^2}\\\\\Rightarrow AB= \sqrt{100+400}\\\\\Rightarrow AB= \sqrt{500}\\\\[/tex]
[tex]BC = (-16,12)(-26,-18)\\\\x_1= -16 \ \ y_1=12 \ \ \ and \ \ \ x_2=-26 \ \ y_2=-18\\[/tex]
[tex]\Rightarrow BC= \sqrt{(-26-(-16))^2+(-18-12)^2}\\\\\Rightarrow BC= \sqrt{(-26+16))^2+(-30)^2}\\\\\Rightarrow BC= \sqrt{(-10)^2+(-30)^2}\\\\\Rightarrow BC= \sqrt{100+900}\\\\\Rightarrow BC= \sqrt{1000}\\\\[/tex]
[tex]\Rightarrow {AC^2= AB^2+BC^2}\\\\\Rightarrow (\sqrt{1000})^2= (\sqrt{500+500})^2\\\\\Rightarrow 1000=1000\\[/tex]
The angle is right-angled triangle
