Straight Lines
Centroid and triangle properties
Grade 11
Question:
<p>A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (−1, 1) and (2, 3). Then the centroid of this triangle is</p>
<p>\(\left(1, \dfrac{7}{3}\right)\)</p>
<p>\(\left(\dfrac{1}{3}, 2\right)\)</p>
<p>\(\left(\dfrac{1}{3}, 1\right)\)</p>
<p>\(\left(\dfrac{1}{3}, \dfrac{5}{3}\right)\)</p>
Step-by-Step Solution
Key Concept: If a vertex is V and midpoints of two sides through V are M₁ and M₂, then the other two vertices can be found using the property that the midpoint formula gives us B = 2M₁ - V and C = 2M₂ - V. The centroid is then the average of all three vertices.
<p><strong>Step 1:</strong> Let A = (1, 2) be the given vertex. Let M₁ = (−1, 1) and M₂ = (2, 3) be the midpoints of sides AB and AC respectively.</p><p><strong>Step 2:</strong> Since M₁ is the midpoint of AB: M₁ = ((1 + x_B)/2, (2 + y_B)/2) = (−1, 1)<br>Solving: (1 + x_B)/2 = −1 ⟹ x_B = −3<br>(2 + y_B)/2 = 1 ⟹ y_B = 0<br>Therefore B = (−3, 0)</p><p><strong>Step 3:</strong> Since M₂ is the midpoint of AC: M₂ = ((1 + x_C)/2, (2 + y_C)/2) = (2, 3)<br>Solving: (1 + x_C)/2 = 2 ⟹ x_C = 3<br>(2 + y_C)/2 = 3 ⟹ y_C = 4<br>Therefore C = (3, 4)</p><p><strong>Step 4:</strong> The centroid G of triangle ABC is: G = ((x_A + x_B + x_C)/3, (y_A + y_B + y_C)/3)<br>G = ((1 − 3 + 3)/3, (2 + 0 + 4)/3) = (1/3, 2)</p><p>∴ Answer: (1/3, 2)</p>
Correct Answer: A