The vertices of a triangle are A(−1, 3), B(−2, 2) and 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:

The sides are $$AB: x-y+4=0$$, $$BC: 3x+5y-4=0$$, and $$AC: x+y-2=0$$.

Shifting a line $$ax+by+c=0$$ inward by $$d$$ units: $$\frac{ax+by+c}{\sqrt{a^2+b^2}} = \pm d$$.

For $$AC (x+y-2=0)$$, the distance from origin is $$\frac{|-2|}{\sqrt{2}} = \sqrt{2}$$. Shifting it 1 unit "inward" (towards the triangle's centroid) results in $$x+y-(2-\sqrt{2})=0$$.

This line is closest to the origin because the original distance was $$\sqrt{2}$$ and we moved 1 unit closer.

Answer: C ($$x+y-(2-\sqrt{2})=0$$)