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.

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 :

(1) \(x-y-(2+\sqrt{2})=0\)

(2) \(-x+y-(2-\sqrt{2})=0\)

(3) \(x+y-(2-\sqrt{2})=0\)

(4) \(x+y+(2-\sqrt{2})=0\)

Answer 1

Correct option is (3) \(x+y-(2-\sqrt{2})=0\)

Equation of \(\mathrm{AC} \rightarrow \mathrm{x}+\mathrm{y}=2\)

Equation of line parallel to \(\mathrm{AC} \mathrm{x}+\mathrm{y}=\mathrm{d}\)

\(\left|\frac{\mathrm{d}-2}{\sqrt{2}}\right|=1\)

\(\mathrm{d}=2-\sqrt{2}\)

Equation of new required line

\(\mathrm{x}+\mathrm{y}=2-\sqrt{2}\)