JEE MAIN - Mathematics (2024 - 4th April Morning Shift - No. 11)
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 :
$$-x+y-(2-\sqrt{2})=0$$
$$x+y-(2-\sqrt{2})=0$$
$$x+y+(2-\sqrt{2})=0$$
$$x-y-(2+\sqrt{2})=0$$
Explanation
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Equation of $$A C: x+y=2$$
Equation of $$A B: x-y+4=0$$
Equation of $$B C: 3 x+5 y=4$$
The line nearest to origin is parallel to $$A C$$ and inward. Let its equation is $$x+y=C$$.
$$\therefore\left|\frac{C-2}{\sqrt{2}}\right|=1$$
$$\therefore \quad C=2-\sqrt{2}$$
$$\therefore$$ required equation line is :
$$x+y-(2-\sqrt{2})=0$$
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