Question:
<p>Let A = (-1, 0) and D = (0, -1). Two points B and C are such that points A, B, C, D are concyclic. Given that AB and CD are parallel and AB has equation x - y + 1 = 0.<br />
If B = (p, 2) and C = (r, 5), then r - s - p is</p>
<p style="display:inline">-1</p>
<p style="display:inline">0</p>
<p style="display:inline">-2</p>
<p style="display:inline">3</p>
Step-by-Step Solution
Key Concept: A cyclic quadrilateral with a pair of parallel sides is an isosceles trapezoid, and if adjacent sides are shown to be perpendicular, the figure must be a rectangle.
<html><body><p><img alt="" data-imgur-src="Uqn9Eax.png" height="136" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/1623761678-kbfpqr.jpg" width="150"/><br/>
Equation of AD : x + y + 1 = 0 ...(i)<br/>
and equation of AB : x - y + 1 = 0 . ..(ii)<br/>
<span class="math-tex">$\Rightarrow$</span> AB <span class="math-tex">$\perp$</span> AD<br/>
Points A, B, C, D concyclic and AB || DC<br/>
<span class="math-tex">$\Rightarrow$</span> Cyclic quadrilateral ABCD is a rectangle<br/>
Point B(p, 2) satisfies (ii) the equation of AB<br/>
<span class="math-tex">$\Rightarrow$</span> p = 1<br/>
Equation of CD : x - y - 1 = 0 ...(iii)<br/>
Point C satisfies (iii)<br/>
<span class="math-tex">$\Rightarrow$</span> r - s = 1 ...(iv)<br/>
Slope of BC = -1<br/>
<span class="math-tex">$\Rightarrow \frac{s-2}{r-1}=-1$</span><br/>
<span class="math-tex">$\Rightarrow$</span> r + s = 3 ...(v)<br/>
Solving (iv) and (v), we get r = 2, s = 1<br/>
Required expression = 2 - 1 - 1 = 0</p></body></html>
Correct Answer: B