Straight Lines
Area of Triangles
Grade 11
Question:
<p><span class="math">A(6, 3)</span>, <span class="math">B(-3, 5)</span>, <span class="math">C(4, -2)</span> and <span class="math">D(x, 3x)</span> are four points. If the area of <span class="math">\triangle DBC</span> and <span class="math">\triangle ABC</span> are in the ratio <span class="math">1:2</span>, then <span class="math">x</span> is equal to</p>
<p>(a) <span class="math">\frac{11}{8}</span></p>
<p>(b) <span class="math">\frac{8}{11}</span></p>
<p>(c) 3</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the area formula for a triangle with coordinates to express both triangle areas, then apply the given ratio condition. The area of a triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃) is ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|.
<p><strong>Step 1: Find the area of triangle ABC.</strong></p><p>Using the coordinate formula with A(6,3), B(-3,5), C(4,-2):</p><p>Area(ABC) = ½|6(5-(-2)) + (-3)((-2)-3) + 4(3-5)|</p><p>= ½|6(7) + (-3)(-5) + 4(-2)|</p><p>= ½|42 + 15 - 8|</p><p>= ½|49| = 49/2</p><p><strong>Step 2: Find the area of triangle DBC.</strong></p><p>With D(x,3x), B(-3,5), C(4,-2):</p><p>Area(DBC) = ½|x(5-(-2)) + (-3)((-2)-3x) + 4(3x-5)|</p><p>= ½|x(7) + (-3)(-2-3x) + 4(3x-5)|</p><p>= ½|7x + 6 + 9x + 12x - 20|</p><p>= ½|28x - 14|</p><p>= ½|28x - 14| = 7|4x - 2|/2</p><p><strong>Step 3: Apply the ratio condition.</strong></p><p>Given: Area(DBC) : Area(ABC) = 1 : 2</p><p>Therefore: Area(DBC)/Area(ABC) = 1/2</p><p>½|28x - 14| / (49/2) = 1/2</p><p>|28x - 14| / 49 = 1/2</p><p>|28x - 14| = 49/2</p><p>28x - 14 = ±49/2</p><p><strong>Step 4: Solve for x.</strong></p><p>Case 1: 28x - 14 = 49/2</p><p>28x = 14 + 49/2 = 28/2 + 49/2 = 77/2</p><p>x = 77/56 = 11/8</p><p>Case 2: 28x - 14 = -49/2</p><p>28x = 14 - 49/2 = 28/2 - 49/2 = -21/2</p><p>x = -21/56 = -3/8</p><p><strong>Step 5: Verify the answer.</strong></p><p>Since x = 11/8 appears in option (a) and matches the expected form, this is our answer.</p><p><strong>∴ Answer: a</strong></p>
Correct Answer: a