Straight Lines
Grade None
Question:
<p>Let B and C be the two points on the line y + x = 0 such that B and C are symmetric with respect to the origin. Suppose A is a point on y - 2x = 2 such that <span class="math-tex">\(\triangle\)</span>ABC is an equilateral triangle. Then, the area of the <span class="math-tex">\(\triangle\)</span>ABC is</p>
<p style="display:inline"><span class="math-tex">\(2 \sqrt{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{10}{\sqrt{3}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(3 \sqrt{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{8}{\sqrt{3}}\)</span></p>
Step-by-Step Solution
Key Concept: The altitude from vertex A of the equilateral triangle must be the perpendicular bisector of the base BC; since B and C are symmetric about the origin on the line x+y=0, A must be the intersection of the line y-2x=2 and the perpendicular line y=x.
<p>Graph of given line<br />
<img src="https://media-mycbseguide.s3.amazonaws.com/images/question_images/1700044068-22vqhd.jpg" style="height:156px; width:180px" /><br />
At A, x = y and y - 2x = 2<br />
So (x, y) = (-2, -2)<br />
Height from line x + y = 0, h = <span class="math-tex">\(\frac{4}{\sqrt{2}}\)</span><br />
So, area of <span class="math-tex">\(\triangle\)</span> = <span class="math-tex">\(\frac{\sqrt{3}}{4} \frac{h^2}{\sin ^2 60}=\frac{8}{\sqrt{3}}\)</span></p>
Correct Answer: D