Problem 13 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The \(x\)-coordinate 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: [2013] (a) \(2+\sqrt{2}\) (b) \(2-\sqrt{2}\) (c) \(1+\sqrt{2}\) (d) \(1-\sqrt{2}\)

Step-by-Step Solution

Answer

The x-coordinate of the incentre is \(2-\sqrt{2}\).

Step 1: Identify vertices of the triangle

The given points are midpoints of the sides of a triangle. Let's assume the vertices of the triangle to be \((A, B, C)\). Since the midpoints are given as \((0, 1), (1, 1)\), and \((1, 0)\), we can set up equations to find these vertices.

Step 2: Determine vertices

The coordinates of the triangle's vertices can be obtained using the midpoint formula. If the midpoint M of segment AB is \((x_1, y_1)\) then \(A + B = 2x_1\) and \(C + D = 2y_1\).- Let the vertices be: \(P(x_1, y_2)\), \(Q(x_2, y_2)\), \(R(x_3, y_3)\).- Since \((x_1 + x_2)/2 = 1\) and \((y_1 + y_2)/2 = 1\), solve these for vertex points.

Step 3: Solve the system of equations

Given point \((0, 1)\) for the midpoint of vertices A(x1, y1) and B(x2, y2), setup and solve: - For midpoint \((0, 1)\), with equations \((x1+x2)/2 = 0\) and \((y1+y2)/2 =1\), implying \(x1 = -x2\), \(y1=2-y2\).- Similarly, solve for \((1, 1)\) and \((1, 0)\).

Step 4: Calculate the coordinates for vertices P, Q, and R

Using the midpoint equations sequentially, \(A = (-1, 2)\), \(B = (1, 0)\), \(C(1, 2)\).

Step 5: Use incentre formula

The formula for the coordinates of the incenter \((I_x, I_y)\) of the triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) and sides lengths \(a, b, c\) opposite to these vertices is \(I_x = \frac{a x_1 + b x_2 + c x_3}{a+b+c}\) where \(a, b, c\) are the lengths of sides.

Step 6: Calculate side lengths

Calculate lengths of each side of the triangle using distance formula \(AB\), \(BC\), \(CA\). For example, for side \(AB\) where A is \((-1, 2)\) and B is \((1, 0)\), use \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).

Step 7: Find the x-coordinate of the incentre

After calculating sides: \(a, b, c\), use the values in Step 5's formula to solve for \(I_x\): \(-1 \times 1 + \sqrt{2} \times 1 + \sqrt{2} \times 1 / 2\approx1.41\).

Key Concepts

Vertices of Triangle

Understanding the vertices of a triangle is key to solving many problems in coordinate geometry. Vertices are the points where the sides of the triangle meet. When given midpoints of the sides, you can reverse-engineer the positions of the vertices. Suppose you know:

• Midpoint of AB is (0, 1)
• Midpoint of BC is (1, 1)
• Midpoint of CA is (1, 0)

To find the vertices (A, B, C), use the midpoint formula by setting both x-coordinates and y-coordinates separately for each midpoint. For instance, knowing the midpoint formula *(x_1+x_2)/2 = x* and *(y_1+y_2)/2 = y*, you can deduce:

• Point A and B satisfy: \( (x_1+x_2)/2 = 0, (y_1+y_2)/2 = 1 \)
• Similarly, apply for other pairs accordingly.

This reverse calculation allows you to find the original vertices of the triangle given its midpoints.

Midpoint Formula

The midpoint formula is a basic but highly useful tool in coordinate geometry. It helps in finding the exact halfway point between two given points. The formula is as follows: \(M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\), where

• \(x_1, y_1\) are the coordinates of the first endpoint.
• \(x_2, y_2\) are the coordinates of the second endpoint.

Apply this formula when you need to locate the middle point on a side of a triangle formed by its vertices. It simplifies calculations and helps in understanding how different points are related. In the context of the given exercise, using the midpoint formula correctly allowed the identification of the triangle's vertices from provided midpoints.

Incentre Formula

Finding the incentre of a triangle is crucial for many geometric calculations. The incentre is the point where the angle bisectors of a triangle's angles meet, and it's the center of the triangle's inscribed circle (incircle). To find the coordinates of the incentre \(I(x, y)\), you can use the incenter formula:\(I_x = \frac{a \cdot x_1 + b \cdot x_2 + c \cdot x_3}{a + b + c}\)

• \(a, b, c\) are the lengths of the sides opposite to vertices \(,x_1, x_2, x_3\).
• This weighted sum gives you the x-coordinate, with y being similar.

This formula takes into account not just the vertices' locations but their relative distances, which makes it integral to comprehensive triangle analysis.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, allows the application of algebraic methods to solve geometric problems. By placing points on a coordinate plane defined by x and y axes, you can calculate distances, midpoints, and other fundamental elements such as angles and slopes. This geometric framework greatly aids in visualizing problems, providing:

• A systematic way to handle various shapes, like triangles, and calculate properties like area and perimeters.
• Transforms abstract geometric principles into tangible, solvable equations.

Applying coordinate geometry in finding a triangle's incentre combines these calculations, aligning midpoints, vertices, and tangent circles systematically to yield accurate results.