Problem 92 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The abscissa of the incentre of the triangle that has the coordinates of mid points of its sides as \((0,1)(1,1)\) and \((1,0)\) is (A) \(2-\sqrt{2}\) (B) \(1+\sqrt{2}\) (C) \(1-\sqrt{2}\) (D) \(2+\sqrt{2}\)

Step-by-Step Solution

Answer

The correct answer is (A) 2 - √2.

Step 1: Identify Midpoints of Triangle's Sides

The given points are the midpoints of the triangle's sides. Let's denote them as \( M_1(0,1) \), \( M_2(1,1) \), and \( M_3(1,0) \). These are the midpoints of sides \( BC \), \( CA \), and \( AB \) respectively.

Step 2: Apply Midpoint Formula

Let the vertices of the triangle be \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \). Given midpoints mean, \[\left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) = (0,1) \] \[\left( \frac{x_3 + x_1}{2}, \frac{y_3 + y_1}{2} \right) = (1,1) \] \[\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (1,0) \].

Step 3: Solve Midpoint Equations

Solve each equation from the midpoint formulas:1. \( x_2 + x_3 = 0 \) and \( y_2 + y_3 = 2 \)2. \( x_3 + x_1 = 2 \) and \( y_3 + y_1 = 2 \)3. \( x_1 + x_2 = 2 \) and \( y_1 + y_2 = 0 \).

Step 4: Solve for Triangle Vertices

By solving the system of equations, we find:- Solving for \( x \): 1. \( x_2 + x_3 = 0 \) 2. \( x_3 + x_1 = 2 \) 3. \( x_1 + x_2 = 2 \) Adding all, \( 2(x_1 + x_2 + x_3) = 4 \rightarrow x_1 + x_2 + x_3 = 2 \). Substituting, get \( x_2 = -x_3 = 1 \), \( x_1 = 1 \).- Solving for \( y \): 1. \( y_2 + y_3 = 2 \) 2. \( y_3 + y_1 = 2 \) 3. \( y_1 + y_2 = 0 \) Solving gives \( y_1 = 1 \), \( y_2 = -1 \), \( y_3 = 3 \). Thus vertices are \((1,1), (1,-1), (-1,3)\).

Step 5: Use Incentre Formula

For a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\), having lengths \(a\), \(b\), and \(c\), the incenter coordinates are \( ( \, \frac{a x_1 + b x_2 + c x_3}{a+b+c}, \frac{a y_1 + b y_2 + c y_3}{a+b+c} \,) \).

Step 6: Find Side Lengths

Calculate lengths of sides: - \( a = \sqrt{(1 - 1)^2 + (-1 - 3)^2} = \sqrt{16} = 4 \) - \( b = \sqrt{(-1 - 1)^2 + (3 - 1)^2} = \sqrt{8} = 2\sqrt{2} \) - \( c = \sqrt{(-1 - 1)^2 + (3 - 1)^2} = \sqrt{8} = 2\sqrt{2} \).

Step 7: Calculate Incentre's Abscissa

Substitute side lengths into the formula for the x-coordinate of the incenter: \[ x_{incenter} = \frac{4 \cdot 1 + 2\sqrt{2} \cdot 1 + 2\sqrt{2} \cdot (-1)}{4 + 2\sqrt{2} + 2\sqrt{2}} = \frac{4 + 2\sqrt{2} - 2\sqrt{2}}{4 + 4\sqrt{2}} = \frac{4}{4 + 4\sqrt{2}} \] By rationalizing, we find \( x_{incenter} = 2 - \sqrt{2} \).

Key Concepts

Midpoint Formula

Midpoint formula is a fundamental concept in geometry that helps us find the midpoint of a line segment, given the coordinates of its endpoints. This can be exceptionally useful when dealing with triangles and finding relationships between different parts of the triangle.

The formula is given as follows: if you have a line segment with endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \), then the midpoint \( M(x, y) \) is calculated using:

• \( x = \frac{x_1 + x_2}{2} \)
• \( y = \frac{y_1 + y_2}{2} \)

This simple averaging gives us the point that lies exactly halfway along the line, making it a powerful tool for dividing segments and understanding more complex geometric properties.

For example, in the exercise provided, the given points were actually midpoints of the triangle's sides. By applying the midpoint formula, we were able to establish equations that helped us find the vertices of the triangle. This illustrates how the formula can serve as a foundational step for solving more complex geometry problems.

Triangle Vertices

Triangle vertices are crucial for understanding the structure and properties of a triangle. Simply put, the vertices are the three points where the sides of a triangle intersect.

In mathematical problems, determining the coordinates of these vertices is often a key step in solving related geometric exercises. The vertices are typically denoted as \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \).

In the provided problem, we used the known midpoints of the triangle's sides to uncover the vertices. By forming and solving systems of equations from the midpoint formula, we could find each vertex coordinate:

• \( A(1, 1) \)
• \( B(1, -1) \)
• \( C(-1, 3) \)

Solving these equations not only confirms whether the solutions are consistent but also lays the groundwork for subsequent calculations, such as determining the triangle's incenter.

Incenter Formula

The incenter of a triangle is a critical point that represents the center of the circle inscribed within the triangle. This point is equidistant from all three sides of the triangle and offers insight into the triangle's balance and symmetry.

To find the incenter's coordinates, one can use the incenter formula, which requires the side lengths and the coordinates of the vertices. For a triangle with sides of lengths \( a \), \( b \), and \( c \,\) and vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \), the formula is:

• \( x_{incenter} = \frac{a x_1 + b x_2 + c x_3}{a+b+c} \)
• \( y_{incenter} = \frac{a y_1 + b y_2 + c y_3}{a+b+c} \)

In this exercise, we calculated the side lengths using distance formula, and plugged these lengths alongside the vertices' coordinates into the incenter formula to find the x-coordinate of the incenter. Simplifying our result confirmed that \( x_{incenter} = 2 - \sqrt{2} \), revealing not just the incenter's position, but also the balance of the triangle.