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JEE Mains · Maths · STD 11 - 9. straight line

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. A \(x-y-(2+\sqrt{2})=0\)
  2. B  \(-\mathrm{x}+\mathrm{y}-(2-\sqrt{2})=0\)
  3. C \(x+y-(2-\sqrt{2})=0\)
  4. D \(x+y+(2-\sqrt{2})=0\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

equation of \(\mathrm{AC} \rightarrow \mathrm{x}+\mathrm{y}=2\) equation of line parallel to \(\mathrm{AC} \mathrm{x}+\mathrm{y}=\mathrm{d}\) \( \left|\frac{\mathrm{d}-2}{\sqrt{2}}\right|=1 \) \( \mathrm{~d}=2-\sqrt{2}\) \(\mathrm{eq}^{\mathrm{n}}\) of new required line…
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