MathematicsMedium128×since 2002Q5559The vertices of a triangle are A(−1,3),B(−2,2)\mathrm{A}(-1,3), \mathrm{B}(-2,2)A(−1,3),B(−2,2) and C(3,−1)\mathrm{C}(3,-1)C(3,−1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :A−x+y−(2−2)=0-x+y-(2-\sqrt{2})=0−x+y−(2−2)=0Bx+y−(2−2)=0x+y-(2-\sqrt{2})=0x+y−(2−2)=0Cx+y+(2−2)=0x+y+(2-\sqrt{2})=0x+y+(2−2)=0Dx−y−(2+2)=0x-y-(2+\sqrt{2})=0x−y−(2+2)=0Check answerSkip