Straight Lines
Area of Triangles
Grade 11
Question:
<p>Area of a triangle is 5 sq units and two of its vertices are <span class="math">(2, 1)</span> and <span class="math">(3, -2)</span>. If its third vertex is on the line <span class="math">y = x + 3</span>, then it is</p>
<p>(a) <span class="math">\left(\frac{-3}{2}, \frac{3}{2}\right)</span></p>
<p>(b) <span class="math">\left(\frac{13}{2}, \frac{19}{2}\right)</span></p>
<p>(c) Both (a) and (b)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the area formula for a triangle with three vertices: Area = ½|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|. Since the third vertex lies on y = x + 3, substitute this constraint to find the coordinates.
<p><strong>Step 1:</strong> Let the third vertex be P(h, k) where k = h + 3 (since it lies on y = x + 3).</p><p><strong>Step 2:</strong> Use the area formula with vertices A(2, 1), B(3, -2), and P(h, h+3):</p><p>Area = ½|2(-2 - (h+3)) + 3((h+3) - 1) + h(1 - (-2))|</p><p>= ½|2(-5 - h) + 3(h + 2) + 3h|</p><p>= ½|-10 - 2h + 3h + 6 + 3h|</p><p>= ½|4h - 4|</p><p>= 2|h - 1|</p><p><strong>Step 3:</strong> Set the area equal to 5:</p><p>2|h - 1| = 5</p><p>|h - 1| = 5/2</p><p><strong>Step 4:</strong> Solve for h:</p><p>h - 1 = 5/2 → h = 7/2</p><p>OR</p><p>h - 1 = -5/2 → h = -3/2</p><p><strong>Step 5:</strong> Find corresponding k-values using k = h + 3:</p><p>When h = 7/2: k = 7/2 + 3 = 13/2, giving point (7/2, 13/2)</p><p>When h = -3/2: k = -3/2 + 3 = 3/2, giving point (-3/2, 3/2)</p><p><strong>Step 6:</strong> Verify both points produce area = 5 (both satisfy the equation).</p><p><strong>Verification for h = 7/2:</strong> Area = 2|7/2 - 1| = 2(5/2) = 5 ✓</p><p><strong>Verification for h = -3/2:</strong> Area = 2|-3/2 - 1| = 2(5/2) = 5 ✓</p><p><strong>∴ Answer:</strong> c</p>
Correct Answer: c