Problem 28 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the line \(3 x+4 y-24=0\) intersects the \(x\)-axis at the point A and the \(y\)-axis at the point \(B\), then the incentre of the triangle OAB, where \(\mathrm{O}\) is the origin, is: [Jan. \(\mathbf{1 0}, \mathbf{2 0 1 9}\) (I)] (a) \((3,4)\) (b) \((2,2)\) (c) \((4,3)\) (d) \((4,4)\)

Step-by-Step Solution

Answer

The incentre of triangle OAB is at (3, 4).

Step 1: Find Point A

The point A, where the line intersects the x-axis, is found by setting \(y = 0\) in the equation of the line. Substituting \(y = 0\) into \(3x + 4y - 24 = 0\) gives \(3x - 24 = 0\), hence \(x = 8\). Therefore, point A is \((8, 0)\).

Step 2: Find Point B

The point B, where the line intersects the y-axis, is found by setting \(x = 0\). Substituting \(x = 0\) into \(3x + 4y - 24 = 0\) gives \(4y - 24 = 0\), hence \(y = 6\). Therefore, point B is \((0, 6)\).

Step 3: Determine Side Lengths

Calculate the length of the sides OA, OB, and AB. OA, the distance from the origin to A, is 8. OB, the distance from the origin to B, is 6. AB is calculated using the distance formula: \(AB = \sqrt{(8-0)^2 + (0-6)^2} = 10\).

Step 4: Calculate the Incentre Coordinates

The incentre \((x, y)\) of triangle OAB is given by the formula \(\left(\frac{a \cdot x_1 + b \cdot x_2 + c \cdot x_3}{a + b + c}, \frac{a \cdot y_1 + b \cdot y_2 + c \cdot y_3}{a + b + c}\right)\), where \(a\), \(b\), and \(c\) are the lengths of sides opposite to vertices O, A, and B respectively.Using OA=8, OB=6, AB=10, and points O=(0,0), A=(8,0), B=(0,6), we have:\(x = \frac{6 \cdot 8 + 8 \cdot 0 + 10 \cdot 0}{6 + 8 + 10} = 3\)\(y = \frac{6 \cdot 0 + 8 \cdot 0 + 10 \cdot 6}{6 + 8 + 10} = 4\).Thus, the incentre is \((3, 4)\).

Key Concepts

Incentre of a Triangle

The incentre of a triangle is a point inside the triangle which is the center of its incircle. The incircle is a circle that touches all three sides of the triangle at exactly one point on each side. It's a significant point of concurrency in a triangle where the angle bisectors of the triangle meet.
The coordinates of the incentre can be calculated using the weighted average of the vertices of the triangle. If you have a triangle formed by points

• point O at the origin
• point A at \((x_1, y_1)\), and
• point B at \((x_2, y_2)\),

then the formula for the incentre is:\[\left( \frac{a \cdot x_1 + b \cdot x_2 + c \cdot x_3}{a + b + c}, \frac{a \cdot y_1 + b \cdot y_2 + c \cdot y_3}{a + b + c} \right)\]where

• \(a\), \(b\), and \(c\)are the lengths of the sides opposite to vertices \(O\), \(A\), and \(B\), respectively.This formula helps in finding the precise location of the incentre, enabling us to draw the incircle accurately based on the triangle's dimensions and alignment.By using the coordinates and side lengths from the vertex points, you can plug them into the formula to find where the incentre is located within the triangle.

Line Intersection

In coordinate geometry, finding the intersection points where a line meets the axes is fundamental. For a line defined by the equation \(Ax + By + C = 0\):

• The x-intercept is found by setting \(y = 0\) in this equation, and solving for \(x\).
• The y-intercept is found by setting \(x = 0\) and solving for \(y\).

For instance, for the line \(3x + 4y - 24 = 0\), you find:

• The x-intercept at point A by solving \(3x - 24 = 0\), giving \(x = 8\), so A is \((8, 0)\).
• The y-intercept at point B by solving \(4y - 24 = 0\), giving \(y = 6\), so B is \((0, 6)\).

Understanding how to determine these intercepts helps you pinpoint crucial points which play roles in forming geometric shapes like triangles when combined with another point, such as the origin.

Distance Formula

The distance formula is an essential tool in coordinate geometry. It helps us measure the straight line distance between two points in a plane. If you are given two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) between these two points is calculated using the formula:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This formula is derived from the Pythagorean theorem and gives the shortest path between the two points.In the given problem about the triangle OAB:

• We use this formula to calculate \(OA\), \(OB\), and \(AB\).
• For \(AB\), the distance between points A (\(8, 0\))and B (\(0, 6\)) is found as follows:

\[AB = \sqrt{(8 - 0)^2 + (0 - 6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\]This calculation is crucial as it provides the length of the side AB, which is vital for further geometric calculations, like finding the incentre of a triangle, ensuring every part of the triangle is accurately measured and understood.