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

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 abscissa of the incentre is \(2 - \sqrt{2}\) (Option A).

Step 1: Identify Vertices of the Triangle

The given points are the midpoints of the sides of the triangle. Let the vertices of the triangle be \(A\), \(B\), and \(C\). The midpoints are \((0,1)\), \((1,1)\), \((1,0)\). These points suggest the use of the midpoint formula: Between any two points \((x_1, y_1)\) and \((x_2, y_2)\), the midpoint is \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\). We need to solve for the coordinates of \(A\), \(B\), and \(C\).

Step 2: Determine Vertices from Midpoints

For midpoint \((0,1)\), assume \(A = (x_1, y_1)\) and \(B = (x_2, y_2)\), then: \[ \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = (0,1) \]For the \(x\)-coordinate, \(\frac{x_1 + x_2}{2} = 0\) so \(x_1 + x_2 = 0\). For the \(y\)-coordinate, \(\frac{y_1 + y_2}{2} = 1\) so \(y_1 + y_2 = 2\).

Step 3: Calculate Specific Vertex Pairings

Now similarly solve for the other midpoints: - Midpoint \((1,1)\) for vertices \(B\) and \(C\): - \(x_2 + x_3 = 2\), \(y_2 + y_3 = 2\).- Midpoint \((1,0)\) for vertices \(C\) and \(A\): - \(x_3 + x_1 = 2\), \(y_3 + y_1 = 0\).

Step 4: Solve the System of Equations

From the equations:1. \(x_1 + x_2 = 0\)2. \(x_2 + x_3 = 2\)3. \(x_3 + x_1 = 2\)4. \(y_1 + y_2 = 2\)5. \(y_2 + y_3 = 2\)6. \(y_3 + y_1 = 0\)Let's solve them:- From (1), \(x_2 = -x_1\).- Substitute in (2), \(-x_1 + x_3 = 2\) so \(x_3 = 2 + x_1\).- Substitute in (3), \(2 + x_1 + x_1 = 2\) so \(2x_1 = 0\) making \(x_1 = 0\).- Use \(x_1 = 0\) in (1) and (2), \(x_2 = 0\), \(x_3 = 2\).For \(y\)-coordinates:- From (4), \(y_1 + y_2 = 2\).- From (5), \(y_2 + y_3 = 2\).- From (6), \(y_3 + y_1 = 0\).- Solve these to obtain: \(y_1 = 1\), \(y_2 = 1\), \(y_3 = -1\).Thus, vertices are \((0,1), (0,1), (2,-1)\).

Step 5: Determine Incentre Coordinates

The incentre \((x_i, y_i)\) of a triangle can be found using the formula:\[ x_i = \frac{a \cdot x_1 + b \cdot x_2 + c \cdot x_3}{a + b + c} \]where \(a\), \(b\), and \(c\) are the lengths of the sides opposite the vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), respectively.Using the given vertices, calculate the side lengths:- \(a = \sqrt{(0 - 2)^2 + (1 - (-1))^2} = \sqrt{4 + 4} = 2\sqrt{2}\)- \(b = \sqrt{(0-2)^2 + (1-1)^2} = 2\)- \(c = \sqrt{(0-0)^2 + (1-(-1))^2} = 2\)Substitute into the formula:\[ x_i = \frac{2 \sqrt{2} \cdot 0 + 2 \cdot 0 + 2 \cdot 2}{2\sqrt{2} + 2 + 2} \]\[ x_i = \frac{4}{2 + 2+ 2\sqrt{2}} \]

Step 6: Simplify Fraction for Abscissa

Simplify:\[ x_i = \frac{4}{4 + 2\sqrt{2}} = \frac{4}{4(1 + \frac{\sqrt{2}}{2})} \]Rationalize the denominator:\[ x_i = \frac{4}{4 + 2\sqrt{2}} \times \frac{4 - 2\sqrt{2}}{4 - 2\sqrt{2}} \]\[ x_i = \frac{16 - 8\sqrt{2}}{16 - 8} = \frac{16 - 8\sqrt{2}}{8} \]\[ x_i = 2 - \sqrt{2} \]This corresponds to option (A).

Key Concepts

Midpoint Formula

The midpoint formula is a simple yet powerful tool in geometry that helps find the center point between two given points. Suppose you have two points with coordinates

• \((x_1, y_1)\)
• \((x_2, y_2)\)

, then the midpoint

• \(M\)

of the line segment connecting these points is given by:\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]This can be especially useful when trying to find the center of a segment for equations, like in this problem where the midpoints of the sides of a triangle are given. By using the midpoint formula, you can determine the vertices of the triangle, which is a crucial step in finding the incenter. You use this by setting each midpoint’s formula equal to the given midpoint coordinates to solve for the unknown variables.

System of Equations

Solving a system of equations can reveal unknown values for multiple variables that relate to each other through several linear relationships. In this exercise, after using the midpoint formula, you quickly get numerous equations concerning the unknown vertices of the triangle. Given:

• \(x_1 + x_2 = 0\)
• \(x_2 + x_3 = 2\)
• \(x_3 + x_1 = 2\)
• \(y_1 + y_2 = 2\)
• \(y_2 + y_3 = 2\)
• \(y_3 + y_1 = 0\)

To solve these, take one equation at a time. Start by solving the simplest ones and substitute the values into others to find the new variables in a systematic approach. We can say:- From \(x_1 + x_2 = 0\), we get \(x_2 = -x_1\)Using the above, substitute \(x_2 = -x_1\) in \(x_2 + x_3 = 2\) to find \(x_3\).Apply the same logic for the \(y\)-coordinates. Solving such systems gives exact positions of each triangle vertex. These series of substitutions are essential for determining every variable needed from the set equations.

Triangle Vertices Calculation

Calculating the vertices of a triangle when given midpoints requires reversing the midpoint formula. Each midpoint gives rise to two equations, allowing you to calculate the coordinates of the vertices of the triangle. For this task, solve: 1. Use

• \((0, 1)\) with assumed vertices \((A, B)\)

to create two linear equations.2. Use

• \((1, 1)\) for vertices \((B, C)\)
• \((1, 0)\) for vertices \((C, A)\)

applying similar processes.These calculations involve:

• Finding pairs like \((y_1, y_2)\) where you know the combination equals a midpoint value.
• Solve for \((x_1, y_1), (x_2, y_2)\) by using solved variables from another equation.

In this example, completing this part of the exercise gives exact positions for\(A, B, C\), which are:

• \((0,1), (0,1),(2,-1)\)

With these precise points, you can easily move forward to find the coordinate of the incenter, which is another key step in many geometrical problems.