Question

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 :

• A. \( x - y - \left( 2 + \sqrt{2} \right) = 0 \)
• B. \( -x + y - \left( 2 - \sqrt{2} \right) = 0 \)
• C. \( x + y - \left( 2 - \sqrt{2} \right) = 0 \)
• D. \( x + y + \left( 2 + \sqrt{2} \right) = 0 \)

Answer 1

The Correct Option is C

Given points are \(A(-1, 3)\) and \(C(3, -1)\). 

The slope of line \(AC\) is given by 
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 3}{3 + 1} = -1. \] 
Using the point-slope form equation for line \(AC\) passing through point \(A(-1, 3)\):
\[ y - 3 = -1(x + 1) \implies x + y = 2. \] 
To find the equation of a line parallel to \(AC\) but shifted inward by a unit distance, we utilize the formula for the distance between parallel lines. 

For a line of the form \(ax + by + c = 0\), a parallel line shifted by a perpendicular distance \(d\) is given by modifying the constant term:
\[ |d| = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}. \] 
For the given line \(x + y = 2\), 

the equation of a parallel line shifted inward by a distance \(\sqrt{2}\) becomes:
\[ x + y = 2 - \sqrt{2}. \]