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

The coordinates of points \(A\) and \(C\) are \(A(-1, 3)\) and \(C(3, -1)\), respectively.

The slope of the line segment \(AC\) is calculated as:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 3}{3 + 1} = -1. \]

Using the point-slope form with point \(A(-1, 3)\), the equation for line \(AC\) is:

\[ y - 3 = -1(x + 1) \implies x + y = 2. \]

To determine the equation of a line parallel to \(AC\) and offset inwards by one unit of distance, the formula for the distance between parallel lines is applied.

For a linear equation in the form \(ax + by + c = 0\), a parallel line displaced by a perpendicular distance \(d\) can be found by adjusting the constant term, using the formula:

\[ |d| = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}. \]

For the line \(x + y = 2\), a parallel line shifted inwards by a distance of \(\sqrt{2}\) has the equation:

\[ x + y = 2 - \sqrt{2}. \]