Straight Lines
Grade None
Question:
<p>The point (2, 1) is translated parallel to the line L : x - y = 4 by <span class="math-tex">\(2 \sqrt{3}\)</span> units. If the new points Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is:</p>
<p style="display:inline">2x + 2y = 1 - <span class="math-tex">\(\sqrt 6\)</span></p>
<p style="display:inline">x + y = 3 - <span class="math-tex">\(2\sqrt 6\)</span></p>
<p style="display:inline">x + y = 2 - <span class="math-tex">\(\sqrt 6\)</span></p>
<p style="display:inline">x + y = 3 - <span class="math-tex">\(3\sqrt 6\)</span></p>
Step-by-Step Solution
Key Concept: Translate the point using parametric coordinates based on the line's slope and distance, then apply the perpendicular condition and quadrant constraints.
<html><body><p>x - y = 4<br/>
To find equation of R<br/>
slope of L = 0 is 1<br/>
<span class="math-tex">$\Rightarrow$</span> slope of QR = -1<br/>
Let QR isy = mx + c<br/>
y = -x + c<br/>
x + y - c - 0<br/>
distance of QR from (2, 1) is <span class="math-tex">$2 \sqrt{3}$</span><br/>
<span class="math-tex">$2 \sqrt{3}$</span> = <span class="math-tex">$\frac{|2+1-c|}{\sqrt{2}}$</span><br/>
<img alt="" data-imgur-src="1PxYw9G.png" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/1621087843-e5s498.jpg" style="width: 200px; height: 149px;"/><br/>
<span class="math-tex">$2 \sqrt{6}=|3-c|$</span><br/>
c - 3 = <span class="math-tex">$\pm 2 \sqrt{6}$</span> c = 3 <span class="math-tex">$\pm 2 \sqrt{6}$</span><br/>
Line can be x + y = 3 <span class="math-tex">$\pm 2 \sqrt{6}$</span><br/>
x + y = 3 - <span class="math-tex">$2 \sqrt{6}$</span></p></body></html>
Correct Answer: B