TS EAMCET 4 May 2018 Shift 2 Solved Paper

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, $AB=\sqrt{9+1}$ $=\sqrt{10}$ And, $BC=\sqrt{32}$ $=4\sqrt{2}$ And, $CA=\sqrt{10}$ This implies, $AB=CA$ So, the triangle $ABC$ is an isosceles triangle.