Question

Classify ΔABC with vertices A(-1, 1), B(1, 3), and C(3, -1).

a. isosceles acute 
b. scalene acute 
c. isosceles right 
d. equilateral

Answer 1

Isosceles acute, answer a.

Here's the logic:

AC^2=(1-(-1))^2+(-1-3)^2=4+16=20.

BC^2=(-1-3)^2+(1-3)^2=16+4=20. Therefore AC=BC, isosceles.

Angle C=tan^-1((3-(-1))/(3-1))-tan^-1((1-(-1))/(3-(-1))=tan^-1(2)-tan^-1(1/2)=36.87° approx, which is acute.

(The angle C is measured in relation to the triangle. Because AC and BC have negative slope the angle between them at C is also negative, but for the purpose of establishing angle C in relation to the triangle rather than to the reference frame (x-y coordinate system), only the magnitudes of the relevant angles have been considered.)

Answer 2

A. isosceles acute

 

XD