Problem 16 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If a vertex of a triangle is \((1,1)\) and the mid points of two sides through this vertex are \((-1,2)\) and \((3,2)\) then the centroid of the triangle is (a) \(\left(-1, \frac{7}{3}\right)\) (c) \(\left(1, \frac{7}{3}\right)\) (b) \(\left(\frac{-1}{3}, \frac{7}{3}\right)\) (d) \(\left(\frac{1}{3}, \frac{7}{3}\right)\)

Step-by-Step Solution

Step 1: Identify Triangle Points

The problem provides the vertex of the triangle at \((1,1)\) and the midpoints on two sides: \((-1,2)\) and \((3,2)\). We must now identify the remaining vertices \((x_1, y_1)\) and \((x_2, y_2)\) of the triangle.

Step 2: Use Midpoint Formulas

Using the midpoint formula \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) for both midpoints, set up the equations:1. For midpoint \((-1,2)\), the equation is \(\left(\frac{1+x_2}{2}, \frac{1+y_2}{2}\right) = (-1,2)\).2. For midpoint \((3,2)\), the equation is \(\left(\frac{1+x_1}{2}, \frac{1+y_1}{2}\right) = (3,2)\).

Step 3: Solve for Additional Vertices

Solve the equations from the midpoints:- From the first equation \(\left(\frac{1+x_2}{2} = -1, \frac{1+y_2}{2} = 2\right)\), we get \(x_2 = -3\) and \(y_2 = 3\).- From the second equation \(\left(\frac{1+x_1}{2} = 3, \frac{1+y_1}{2} = 2\right)\), we find \(x_1 = 5\) and \(y_1 = 3\).

Key Concepts

Vertex coordinates

In the context of triangle geometry, identifying vertex coordinates is essential. A vertex is a point where two sides of a triangle meet. Each vertex can be represented by its coordinate pair

• For triangle vertices, each has coordinates typically given as
  • \( (x_1, y_1) \)
  • \( (x_2, y_2) \)
  • \( (x_3, y_3) \)

This pair gives its location on an

• x-y coordinate plane.

In our problem, one vertex is provided as

• \((1,1)\).

Knowing these coordinates is the first step in solving problems related to triangle geometry. This is because they provide fixed points from which we calculate other properties, such as centroids or midpoints.

Midpoint formula

The midpoint formula is a key mathematical tool in geometry used to find the middle point between two given points. This formula is particularly useful when dealing with triangles as it aids in tasks like finding midpoints of sides. The formula is written as:\[\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)\]Applying this formula helps us find the coordinates of a midpoint by taking the average of the x-coordinates and y-coordinates, respectively. In our case, two midpoints are provided:

• \((-1,2)\)
• \((3,2)\)

These midpoints result from the coordinates of the triangle’s vertices. The calculated values are crucial for deducing unknown vertex coordinates or other properties of the triangle, including the centroid. By equating these midpoints to their corresponding formulas, we can derive other necessary coordinates needed for further calculations.

Triangle geometry

Within triangle geometry, understanding the concept of centroids is fundamental. The centroid is the point where the three medians of a triangle intersect, which also represents the center of mass or balance point of the triangle. To compute the coordinates of the centroid

• \((G_x, G_y)\)

of a triangle with vertices:

• \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\)

we use the formula\[\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)\]In the given exercise, once we have determined all vertex coordinates, applying the formula gives us the centroid's exact location. This understanding consolidates the relationship between vertices and triangle stability, depicting how each point's position influences the centroid directly. Harnessing this geometric property provides deeper insights into the configurations and properties of triangles.