The vertices of a triangle are $\mathrm{A}(-1,3), \mathrm{B}(-2,2)$ and $\mathrm{C}(3,-1)$. A new triangle…

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

Solution


equation of $\mathrm{AC} \rightarrow \mathrm{x}+\mathrm{y}=2$ equation of line parallel to $\mathrm{AC} \mathrm{x}+\mathrm{y}=\mathrm{d}$ $\begin{aligned} & \left|\frac{\mathrm{d}-2}{\sqrt{2}}\right|=1 \\ & \mathrm{~d}=2-\sqrt{2} \end{aligned}$ $\mathrm{eq}^{\mathrm{n}}$ of new required line $\mathrm{x}+\mathrm{y}=2-\sqrt{2}$

Asked in: JEE Main 2024 (04 Apr Shift 1)

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