Problem 5 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

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 : \(\quad\) [April 12, 2019 (II)] (a) \(\left(1, \frac{7}{3}\right)\) (b) \(\left(\frac{1}{3}, 2\right)\) (c) \(\left(\frac{1}{3}, 1\right)\) (d) \(\left(\frac{1}{3}, \frac{5}{3}\right)\)

Step-by-Step Solution

Answer

The centroid of the triangle is \(\left(\frac{1}{3}, 2\right)\), option (b).

Step 1: Identify the vertices

The given triangle has a vertex at \((1,2)\) (let's call it \(A\)). The midpoints of sides through this vertex are \((-1,1)\) and \((2,3)\). Since a midpoint is the average of the endpoints, we use these points to find the locations of the other two vertices of the triangle.

Step 2: Calculate Coordinates of Other Vertices

Since \((-1,1)\) is the midpoint of a side through \(A\), we use the formula for the midpoint to find the endpoint of that side: \(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\). For one side, \((-1,1) = \left(\frac{1 + x}{2}, \frac{2 + y}{2}\right)\), solving gives \(x = -3\) and \(y = 0\), so one vertex is \((-3,0)\). For the other side, \((2,3) = \left(\frac{1 + x}{2}, \frac{2 + y}{2}\right)\), solving gives \(x = 3\) and \(y = 4\), so the other vertex is \((3,4)\).

Step 3: Calculate the Centroid

The centroid \(G\) of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is given by the formula \(G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)\). Substituting the vertices \((1, 2), (-3, 0), (3, 4)\), we calculate \(G = \left(\frac{1 - 3 + 3}{3}, \frac{2 + 0 + 4}{3}\right) = \left(\frac{1}{3}, \frac{6}{3}\right) = \left(\frac{1}{3}, 2\right)\).

Key Concepts

Midpoint Formula

Understanding the midpoint formula is essential when dealing with triangles and their vertices in coordinate geometry. The midpoint of a line segment is the point that is exactly halfway between the endpoints. To find this point in the coordinate plane, we utilize the midpoint formula:

\[M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

In this formula, \(x_1\), \(y_1\) and \(x_2\), \(y_2\) are the coordinates of the endpoints of the line segment. This is essentially about finding the average of the x-coordinates and the y-coordinates of these endpoints.

In our example, the midpoint \((-1,1)\) tells us what the coordinates of each end of one line segment are. By plugging into the formula, we could solve for the unknown vertex. It helps us confirm the properties of symmetry in a triangle and verify if our calculated vertices are correct.

Centroid Formula

The centroid of a triangle is commonly referred to as its "center of mass" or "barycenter." It is the point where all three medians intersect and serves as a balancing point for the triangle. The centroid formula is straightforward:

\[G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)\]

Each \(x\) and \(y\) represents the coordinate of one of the triangle's vertices. The formula provides the coordinates of this balance point by averaging the x and y coordinates of all vertices.

In the problem at hand, after identifying the vertices as \(1, 2\), \(-3, 0\), and \(3, 4\), we apply the centroid formula and find it results in the point \(\left( \frac{1}{3}, 2 \right)\). This verification reinforces our calculation of each vertex using the midpoint formula, showcasing the interconnectedness of these geometric tools.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. This branch of mathematics allows us to interpret geometric shapes and their properties using algebra.

Applying this technique to triangles means using the coordinates of points, such as vertices or midpoints, to solve various problems like finding the centroid. This combines geometric principles with algebraic equations.

We relied on coordinate geometry throughout this exercise by using known midpoint coordinates to uncover hidden vertices and by using these vertex coordinates within formulas to deduce the centroid. This emphasizes its power in transforming geometric concepts into understandable and computable solutions.

• It allows a bridge between algebra and geometry.
• The process helps solve for unknown points and verify already known ones.
• It offers precise solutions to problems that might be visually complex or intricate.