Question:

To find the equation of the side of the new triangle nearest to the origin after shifting the sides of the original triangle by one unit inwards, we follow these steps: Step 1: First, let's find the equations of the lines representing the sides of the original triangle. The vertices of the triangle are $A(-1,3)$, $B(-2,2)$, and $C(3,-1)$. Step 2: The equation of the line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ can be found using the formula for the slope of a line: $m = \frac{y_2 - y_1}{x_2 - x_1}$. Then, using the point-slope form of the line equation, $y - y_1 = m(x - x_1)$, we can find the equation of the line. Step 3: Let's calculate the equation of the line for side AC. The slope $m_{AC}$ is given by $\frac{-1 - 3}{3 - (-1)} = \frac{-4}{4} = -1$. Using point-slope form with point $A(-1,3)$, we have: $$ \begin{aligned} y - 3 &= -1(x - (-1)) \\ y - 3 &= -x - 1 \\ y &= -x + 2 \end{aligned} $$ So, the equation of the line AC is $y = -x + 2$ or $x + y - 2 = 0$. Step 4: To shift this line one unit inwards, we need to find the line that is parallel to $x + y - 2 = 0$ and one unit closer to the origin. The general equation of a line parallel to $x + y - 2 = 0$ is $x + y + c = 0$. The distance between two parallel lines $x + y + c_1 = 0$ and $x + y + c_2 = 0$ is given by $\frac{|c_2 - c_1|}{\sqrt{1^2 + 1^2}} = \frac{|c_2 - c_1|}{\sqrt{2}}$. We want this distance to be 1, so $\frac{|c_2 - c_1|}{\sqrt{2}} = 1$, which implies $|c_2 - c_1| = \sqrt{2}$. Since $c_1 = -2$, $c_2$ should be either $-2 + \sqrt{2}$ or $-2 - \sqrt{2}$ to be closer to the origin. Step 5: Considering the line $x + y - 2 = 0$ and shifting it inwards by 1 unit, the new line equation will be $x + y - (2 - \sqrt{2}) = 0$ because we are looking for the line that is closer to the origin, and $2 - \sqrt{2}$ is the value that makes the line closer to the origin compared to $2 + \sqrt{2}$. Step 6: Therefore, the equation of the side of the new triangle nearest to the origin is $x + y - (2 - \sqrt{2}) = 0$, which matches option 3. Therefore: $x+y-(2-\sqrt{2})=0$ <div class="key-concept"><strong>Key Concept:</strong> Side $AC$: passes through $(-1,3)$ and $(3,-1)$, slope $=-1$, equation $x+y=2$. The inward parallel line is $x+y=d$ where $|d-2|/\sqrt{2}=1\Rightarrow d=2\pm\sqrt{2}$. Inward means closer to interior (which is on the side of $B(-2,2)$: $-2+2=0<2$, so inward means $d<2$, giving $d=2-\sqrt{2}$.</div> <div class="trap-box"><strong>Trap:</strong> Nearest to origin means smallest $|d|/\sqrt{2}$. $d=2-\sqrt{2}\approx0.59$, so line is $x+y-(2-\sqrt{2})=0$.</div>