JEE Math Practice Question
The vertices of a triangle are $$\mathrm{A}(-1,3), \mathrm{B}(-2,2)$$ and $$\mathrm{C}(3,-1)$$. 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.$$-x+y-(2-\sqrt{2})=0$$
- B.$$x+y-(2-\sqrt{2})=0$$
- C.$$x+y+(2-\sqrt{2})=0$$
- D.$$x-y-(2+\sqrt{2})=0$$
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