Problem 87 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{P}=(-1,0), \mathrm{Q}=(0,0)\) and \(\mathrm{R}=(3,3 \sqrt{3})\) be three point. The equation of the bisector of the angle PQR is [2007] (a) \(\frac{\sqrt{3}}{2} x+y=0\) (b) \(x+\sqrt{3 y}=0\) (c) \(\sqrt{3} x+y=0\) (d) \(x+\frac{\sqrt{3}}{2} y=0\)

Step-by-Step Solution

Answer

The equation of the bisector of the angle PQR is (c) \(\sqrt{3}x + y = 0\).

Step 1: Find Vectors

First, determine the vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{QR} \). \( \overrightarrow{PQ} = (0 - (-1), 0 - 0) = (1, 0) \) and \( \overrightarrow{QR} = (3 - 0, 3\sqrt{3} - 0) = (3, 3\sqrt{3}) \).

Step 2: Find Unit Vectors

Calculate the magnitudes of \( \overrightarrow{PQ} \) and \( \overrightarrow{QR} \): \( |\overrightarrow{PQ}| = \sqrt{1^2 + 0^2} = 1 \) and \( |\overrightarrow{QR}| = \sqrt{3^2 + (3\sqrt{3})^2} = \sqrt{36} = 6 \). Thus, the unit vectors are \( \hat{PQ} = (1, 0) \) and \( \hat{QR} = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \).

Step 3: Calculate Direction of Angle Bisector

The direction of the angle bisector can be found by summing the unit vectors \( \hat{PQ} \) and \( \hat{QR} \): \( \left(1, 0\right) + \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) = \left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right) \).

Step 4: Find Equation of Bisector Line

With the direction vector \( \left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right) \), we can write the equation of the line through point Q (0,0) as \( y = mx \), where \( m = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{\sqrt{3}}{3} \). Hence, the equation is \( y = \frac{\sqrt{3}}{3} x \) or \( \sqrt{3}x - 3y = 0 \). Simplifying gives \( \sqrt{3}x + y = 0 \).

Step 5: Match with Given Options

The derived equation \( \sqrt{3}x + y = 0 \) matches option (c). Thus, the correct answer is option (c).

Key Concepts

Equation of a Line

When you're dealing with lines in coordinate geometry, understanding the equation of a line is key. The most common form of a line equation is the slope-intercept form: \( y = mx + b \). Here, \( m \) is the slope of the line, and \( b \) is the y-intercept, which is where the line crosses the y-axis.
For example, if a line passes through the origin, that means \( b = 0 \) and the line's equation is simply \( y = mx \). This slope \( m \) can be calculated by the change in \( y \) over the change in \( x \) between two points on the line, or \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
Lines can also be represented in different forms, such as the point-slope form \( y - y_1 = m(x - x_1) \), if a point \((x_1, y_1)\) on the line is known, or in general form \( Ax + By + C = 0 \). Each form can be useful depending on the available information.

Angle Bisector

An angle bisector is a line or ray that divides an angle into two equal parts. In coordinate geometry, this often involves using the concept of vectors to find the directional components. The direction of the angle bisector in question is found by adding unit vectors along the lines forming the angle.
In our problem, we have two vectors \( \overrightarrow{PQ} \) and \( \overrightarrow{QR} \). By converting these vectors into unit vectors and adding them, we determine the direction of the line bisecting the angle \( \angle PQR \). This bisector is important in tasks involving symmetry and minimization of distances between geometric figures.

Vectors

Vectors are fundamental in geometry. They represent both direction and magnitude. Consider a vector \( \overrightarrow{v} = (x, y) \). Here, \( x \) and \( y \) are the components, showing the movement along the respective axes.
Vectors are particularly useful for determining directions and distances between points. The length or magnitude of a vector \( \overrightarrow{v} \) is given by the formula \( |\overrightarrow{v}| = \sqrt{x^2 + y^2} \).
Adding vectors involves summing their corresponding components: \( \overrightarrow{u} + \overrightarrow{v} = (x_1 + x_2, y_1 + y_2) \). This operation can help in finding directions, such as when determining an angle bisector.

Unit Vector

A unit vector is a vector that has a magnitude of 1 and is used to denote direction. If you have a vector \( \overrightarrow{v} = (x, y) \), you can convert it to a unit vector by dividing each component by the length of the vector.
This is calculated as \( \hat{v} = \left(\frac{x}{|\overrightarrow{v}|}, \frac{y}{|\overrightarrow{v}|}\right) \), where \( |\overrightarrow{v}| = \sqrt{x^2 + y^2} \).
Unit vectors are crucial because they offer a clear direction without scaling the vector's effect, particularly useful in operations such as finding angles and determining directional vectors for lines and bisectors in geometry.