Question:
<p>Two sides of a rhombus are along the lines, x - y + 1 = 0 and 7x - y - 5 = 0. If its diagonals intersect at (-1, -2), then which one of the following is a vertex of this rhombus?</p>
<p style="display:inline">(-3, -9)</p>
<p style="display:inline"><span class="math-tex">\(\left(\frac{1}{3},-\frac{8}{3}\right)\)</span></p>
<p style="display:inline"><span class="math-tex">\(\left(-\frac{10}{3},-\frac{7}{3}\right)\)</span></p>
<p style="display:inline">(-3, -8)</p>
Step-by-Step Solution
Key Concept: The intersection of diagonals in a rhombus is the center of symmetry, which lies exactly midway between the two pairs of opposite parallel sides.
<p><img alt="" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/1621087840-srrx3b.jpg" style="height:164px; width:200px" /><br />
Let other two sides of rhombus are<br />
x - y + <span class="math-tex">\(\lambda\)</span> = 0 and 7x - y + <span class="math-tex">\(\mu\)</span> = 0<br />
then O is equidistant from AB and DC and from AD and BC<br />
<span class="math-tex">\(\therefore|-1+2+1|\)</span> = <span class="math-tex">\(|-1+2+\lambda|\)</span> <span class="math-tex">\(\Rightarrow\)</span> <span class="math-tex">\(\lambda\)</span> = -3<br />
and <span class="math-tex">\(|-7+2-5|\)</span> = <span class="math-tex">\(|-7+2+\mu|\)</span> <span class="math-tex">\(\Rightarrow\)</span> <span class="math-tex">\(\mu\)</span> = 15<br />
<span class="math-tex">\(\therefore\)</span> Other two sides are x - y - 3 = 0 and 7x - y + 15 = 0<br />
<span class="math-tex">\(\therefore\)</span> On solving the equations. of sides pairwise, we get the vertices as<br />
<span class="math-tex">\(\left(\frac{1}{3}, \frac{-8}{3}\right)\)</span>, (1, 2), <span class="math-tex">\(\left(\frac{-7}{3}, \frac{-4}{3}\right)\)</span>, (-3, -6)</p>
Correct Answer: B