Consider the straight line equation, x+3y−4=0,x+y−4=0 and 3x+y−4=0 The triangle formed will be,
So, x+2y−4=0 3x+y−4=0 3x+9y−12=0 For the point A intersection, 3x+y−4=0 −‌‌−‌+ –––––––––––– 8y=8 y=1 This implies, 3x+1−4=0 3x=3 x=1 For the intersection point B, 3x+y−4=0 x+y−4=0 −‌‌−‌+ –––––––––––– 2y=0 y=0 This implies, x=4 Similarly for point C, x=0,y=4 Then, the triangle formed will be,
Then, AB=√9+1 =√10 And, BC=√32 =4√2 And, CA=√10 This implies, AB=CA So, the triangle ABC is an isosceles triangle.