Binomial Theorem
Grade None
Question:
<p>Line <span class="math-tex">\(L_{1}\)</span> of slope 2 and line <span class="math-tex">\(L_{2}\)</span> of slope <span class="math-tex">\(\frac{1}{2}\)</span> intersect at the origin <span class="math-tex">\(O\)</span>. In the first quadrant, <span class="math-tex">\(P_{1}, P_{2}, \ldots ., P_{12}\)</span> are 12 points on line <span class="math-tex">\(L_{1}\)</span> and <span class="math-tex">\(Q_{1}\)</span>, <span class="math-tex">\(Q_{2}, \ldots ., Q_{9}\)</span> are 9 points on line <span class="math-tex">\(L_{2}\)</span>. Then the total number of triangles, that can be formed having vertices at three of the 22 points <span class="math-tex">\(O, P_{1}\)</span>, <span class="math-tex">\(P_{2}, \ldots, P_{12}, Q_{1}, Q_{2}, \ldots ., Q_{9}\)</span>, is:</p>
<p style="display:inline">1134</p>
<p style="display:inline">1188</p>
<p style="display:inline">1080</p>
<p style="display:inline">1026</p>
Step-by-Step Solution
Key Concept: To form a triangle from points on two intersecting lines, select three points such that they do not all lie on the same line by combining points from both lines.
<p>1 point from <span class="math-tex">$L_{1}$</span> and 2 from <span class="math-tex">$L_{2}$</span> :<br />
<span class="math-tex">$12 \times{ }^{9} C_{2}=432$</span><br />
2 points from L1 and 1 from L2:<br />
<span class="math-tex">${ }^{12} C_{2} \times 9=594$</span><br />
1 point from <span class="math-tex">$L_{1}, 1$</span> from <span class="math-tex">$L_{2}$</span>, and <span class="math-tex">$O$</span> :<br />
<span class="math-tex">$12 \times 9=108$</span><br />
Total: <span class="math-tex">$432+594+108=1134$</span></p>
Correct Answer: A