Problem 78 - A Complete Resource Book in Mathematics for JEE Main

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) \(\left(-1, \frac{7}{3}\right)\) (B) \(\left(\frac{-1}{3}, \frac{7}{3}\right)\) (C) \(\left(1, \frac{7}{3}\right)\) (D) \(\left(\frac{1}{3}, \frac{7}{3}\right)\)

Step-by-Step Solution

Answer

The centroid of the triangle is \((1, \frac{7}{3})\), which is option (C).

Step 1: Identify the Given Coordinates

We know one vertex of the triangle is at \((1,1)\). The midpoints of the segments through this vertex are given as \((-1,2)\) and \((3,2)\).

Step 2: Use Midpoint Formula

The midpoint formula for a segment between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\). For each midpoint, set up equations to find the unknown vertices.

Step 3: Solve for Other Vertices

1. Let the coordinates of the second vertex be \((x, y)\). - Since \((-1,2)\) is the midpoint between \((1,1)\) and \((x,y)\), we have: \[\begin{align*} \frac{1 + x}{2} = -1 & \Rightarrow x = -3 \ \frac{1 + y}{2} = 2 & \Rightarrow y = 3 \end{align*}\] - So, the second vertex is \((-3,3)\).2. Let the coordinates of the third vertex be \((a, b)\). - Since \((3,2)\) is the midpoint between \((1,1)\) and \((a,b)\), we have: \[\begin{align*} \frac{1 + a}{2} = 3 & \Rightarrow a = 5 \ \frac{1 + b}{2} = 2 & \Rightarrow b = 3 \end{align*}\] - So, the third vertex is \((5,3)\).

Step 4: Calculate the Centroid

The centroid \((G)\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]Substitute the vertices \((1,1)\), \((-3,3)\), and \((5,3)\):\[ G = \left( \frac{1 - 3 + 5}{3}, \frac{1 + 3 + 3}{3} \right) = \left( \frac{3}{3}, \frac{7}{3} \right) = (1, \frac{7}{3}) \]

Step 5: Select the Correct Option

From the centroid calculations, we have \( G = (1, \frac{7}{3}) \), which corresponds to option (C).

Key Concepts

Midpoint Formula

The midpoint formula is a vital tool in coordinate geometry, especially when dealing with triangles and other geometric figures. It allows us to find the point that exactly halves the distance between two coordinates. Simply put, the formula calculates the average of the x-coordinates and y-coordinates of two points.
This is useful to find the midpoint, which is a point lying halfway along the line segment joining two other points.

• The formula for finding a midpoint is: \[ ext{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
• The x-coordinate of the midpoint is the average of the x-coordinates of the endpoints, and similarly, the y-coordinate is the average of the y-coordinates of the endpoints.
• Using the given points: \((-1,2)\) and \((3,2)\) as midpoints, we reverse apply the formula to find the coordinates of the other vertices.
Remember, finding the middle point helps in transparent calculations and is key to solving several geometric problems efficiently.

Triangle Vertices

In a triangle, vertices are the points where two sides meet. Every triangle has three vertices, and together they determine the triangle's shape and size. In coordinate geometry, these vertices are expressed as ordered pairs: \((x_1, y_1), (x_2, y_2), (x_3, y_3)\).

• In the given exercise, knowing one vertex \((1,1)\) helped us a great deal.
• By using the midpoint coordinates, we could determine the other two vertices of the triangle, namely \((-3,3)\) and \((5,3)\).
• This forms the complete set of vertices required to identify the triangle's shape.
• The connection between vertices determines the types of line segments and angles present in the triangle.

To solve for unknown vertices, start with known vertices and use the midpoint formula as demonstrated. This grants us the full picture of the triangle involved.

Coordinate Geometry

Coordinate geometry is a powerful branch of mathematics where positions of points are defined on a plane using ordered pairs \((x, y)\). It combines algebra and geometry, providing tools to solve geometric problems algebraically.

• The calculation of the centroid is a typical application: \[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
• This formula offers a simple way to find the center of mass, or balancing point, of a triangle formed by its vertices.
• A triangle's centroid is crucial in determining balance points and other properties like symmetry.
• The given problem uses known vertices \((1,1), (-3,3), \) and \((5,3)\) to compute the centroid.

Understanding the fundamentals of coordinate geometry enables students to approach problems involving shapes, areas, and centroids with confidence. By connecting algebra and geometry, they find solutions that are both visual and numerical.