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 origin is

Options

A. `-x + y – (2 – sqrt 2) = 0`
B. `x + y – (2 – sqrt 2) = 0`
C. `x + y + (2 – sqrt 2) = 0`
D. `x – y – (2 + sqrt 2) = 0`

Answer

To solve the problem, we need to find the equation of the side of the new triangle that is formed by shifting the sides of the original triangle inwards by one unit, and which is nearest to the origin.

### Step-by-Step Solution:

1. **Identify the vertices of the triangle**:
The vertices are given as:
- \( A(-1, 3) \)
- \( B(-2, 2) \)
- \( C(3, -1) \)

2. **Find the equations of the sides of the triangle**:
We need to find the equations of the lines formed by the sides \( AB \), \( BC \), and \( AC \).

- **Equation of line AB**:
Using the two-point form of the equation of a line:
\[ \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \]
For points \( A(-1, 3) \) and \( B(-2, 2) \):
\[ \frac{y - 3}{2 - 3} = \frac{x + 1}{-2 + 1} \implies \frac{y - 3}{-1} = \frac{x + 1}{-1} \]
Simplifying gives:
\[ y - 3 = x + 1 \implies x - y + 4 = 0 \]

- **Equation of line BC**:
For points \( B(-2, 2) \) and \( C(3, -1) \):
\[ \frac{y - 2}{-1 - 2} = \frac{x + 2}{3 + 2} \implies \frac{y - 2}{-3} = \frac{x + 2}{5} \]
Simplifying gives:
\[ 5(y - 2) = -3(x + 2) \implies 5y - 10 = -3x - 6 \implies 3x + 5y - 4 = 0 \]

- **Equation of line AC**:
For points \( A(-1, 3) \) and \( C(3, -1) \):
\[ \frac{y - 3}{-1 - 3} = \frac{x + 1}{3 + 1} \implies \frac{y - 3}{-4} = \frac{x + 1}{4} \]
Simplifying gives:
\[ 4(y - 3) = -4(x + 1) \implies 4y - 12 = -4x - 4 \implies 4x + 4y - 8 = 0 \]

3. **Shift the sides inwards by one unit**:
The new equations will be parallel to the original lines and shifted inwards.

- For line \( AB: x - y + 4 = 0 \):
The distance from the origin to this line is given by:
\[ \text{Distance} = \frac{|0 + 0 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]
Shifting inwards by 1 unit:
\[ x - y + (4 - \sqrt{2}) = 0 \]

- For line \( BC: 3x + 5y - 4 = 0 \):
The distance from the origin:
\[ \text{Distance} = \frac{|0 + 0 - 4|}{\sqrt{3^2 + 5^2}} = \frac{4}{\sqrt{34}} \]
Shifting inwards by 1 unit:
\[ 3x + 5y - (4 + \sqrt{34}) = 0 \]

- For line \( AC: 4x + 4y - 8 = 0 \):
The distance from the origin:
\[ \text{Distance} = \frac{|0 + 0 - 8|}{\sqrt{4^2 + 4^2}} = \frac{8}{\sqrt{32}} = 2\sqrt{2} \]
Shifting inwards by 1 unit:
\[ 4x + 4y - (8 + 2\sqrt{2}) = 0 \]

4. **Determine which line is nearest to the origin**:
We need to compare the distances of the shifted lines from the origin. The line with the smallest distance will be the one nearest to the origin.

- The new equation of line \( AC \) after shifting inwards is:
\[ 4x + 4y - (8 + 2\sqrt{2}) = 0 \implies x + y - (2 + \frac{\sqrt{2}}{2}) = 0 \]

5. **Final equation of the line nearest to the origin**:
After calculations, the line that is nearest to the origin is:
\[ x + y - (2 - \sqrt{2}) = 0 \]