The vertices of a triangle are A(-1, 3),B(-2, | vedclass

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(C) The vertices of the triangle are $A(-1, 3)$,$B(-2, 2)$,and $C(3, -1)$.First,find the equations of the sides:Equation of $AC$: The slope $m = \frac

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

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(C) The vertices of the triangle are $A(-1, 3)$,$B(-2, 2)$,and $C(3, -1)$.First,find the equations of the sides:Equation of $AC$: The slope $m = \frac{-1-3}{3-(-1)} = \frac{-4}{4} = -1$. The equation is $y - 3 = -1(x + 1) \Rightarrow x + y = 2$.Equation of $AB$: The slope $m = \frac{2-3}{-2-(-1)} = \frac{-1}{-1} = 1$. The equation is $y - 3 = 1(x + 1) \Rightarrow x - y + 4 = 0$.Equation of $BC$: The slope $m = \frac{-1-2}{3-(-2)} = \frac{-3}{5}$. The equation is $y - 2 = -\frac{3}{5}(x + 2) \Rightarrow 3x + 5y - 4 = 0$.To shift a line $ax + by + c = 0$ inwards by $d=1$ unit,the new line is $ax + by + c \pm \sqrt{a^2+b^2} = 0$.For $AC: x + y - 2 = 0$,the new line is $x + y - 2 \pm \sqrt{2} = 0$. Since the origin $(0,0)$ satisfies $x+y For $AB: x - y + 4 = 0$,the new line is $x - y + 4 \pm \sqrt{2} = 0$. The inward shift is $x - y + 4 - \sqrt{2} = 0$.For $BC: 3x + 5y - 4 = 0$,the new line is $3x + 5y - 4 \pm \sqrt{34} = 0$.The side nearest to the origin is $x + y = 2 - \sqrt{2}$.

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 the origin is:

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