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

Answer: (C)

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

Answer 2

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:
$x + y = 2 - \sqrt{2}.$