Problem 27 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A point \(\mathrm{P}\) moves on the line \(2 x-3 y+4=0\). If \(\mathrm{Q}(1,4)\) and \(\mathrm{R}(3,-2)\) are fixed points, then the locus of the centroid of \(\Delta P Q R\) is a line: (a) with slope \(\frac{3}{2}\) (b) parallel to \(x\)-axis (c) with slope \(\frac{2}{3}\) (d) parallel to \(y\)-axis

Step-by-Step Solution

Answer

The locus is a line with slope \( \frac{2}{3} \) (Option c).

Step 1: Understand the Geometry

We have a triangle \( \triangle PQR \) with point \( P \) moving on a line, and fixed points \( Q(1,4) \) and \( R(3,-2) \). We need to find the locus of the centroid of this triangle.

Step 2: Determine the Equation of the Line

Since point \( P \) lies on the line \( 2x - 3y + 4 = 0 \), any point \( P(x, y) \) satisfies this equation.

Step 3: Recall the Centroid Formula

The centroid \( G \) of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is given by the formula: \[ \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

Step 4: Substitute the Vertices Coordinates

For \( \triangle PQR \), \( x_1 = x \), \( y_1 = y \), \( x_2 = 1 \), \( y_2 = 4 \), \( x_3 = 3 \), \( y_3 = -2 \). The coordinates of the centroid \( G(x_g, y_g) \) are: \[ x_g = \frac{x + 1 + 3}{3} = \frac{x + 4}{3} \]\[ y_g = \frac{y + 4 - 2}{3} = \frac{y + 2}{3} \]

Step 5: Express y in terms of x

Since \( P(x, y) \) lies on \( 2x - 3y + 4 = 0 \), solve for y:\[ 3y = 2x + 4 \]\[ y = \frac{2}{3}x + \frac{4}{3} \]

Step 6: Substitute into the y-coordinate of the Centroid

Substitute \( y = \frac{2}{3}x + \frac{4}{3} \) into the expression for \( y_g \):\[ y_g = \frac{\left(\frac{2}{3}x + \frac{4}{3}\right) + 2}{3} = \frac{2x + 4 + 6}{9} = \frac{2x + 10}{9} \]

Step 7: Derive the Equation of the Locus

Since \( x_g = \frac{x + 4}{3} \) and \( y_g = \frac{2x + 10}{9} \), relate them:\[ y_g = \frac{2}{3}x_g + \frac{2}{3} \]This is the equation of the line in the form \( y = mx + c \) where \( m = \frac{2}{3} \) and \( c = \frac{2}{3} \).

Step 8: Conclusion

The locus of the centroid is a line with slope \( \frac{2}{3} \). Therefore, the correct answer is (c) with slope \( \frac{2}{3} \).

Key Concepts

triangle geometry

In geometry, triangles are fundamental shapes consisting of three sides and three vertices. It's important to understand that a triangle can be uniquely defined by its vertices, which are simply the points at which two sides meet. In the problem at hand, we have a triangle denoted as \( \triangle PQR \), where \( Q(1,4) \) and \( R(3,-2) \) are fixed points, and \( P \) moves along a given line. Understanding the positioning and movement of these vertices is key to solving problems about paths traced by points associated with the triangle, such as centroids.

• Vertices: The points \( P, Q, \) and \( R \).
• Sides: The line segments connecting these vertices, such as \( PQ, QR, \) and \( RP \).

As \( P \) moves, the shape of \( \triangle PQR \) changes, influencing other geometric properties like the location of its centroid.

centroid formula

A centroid of a triangle is the point where its three medians intersect. The centroid is often considered the triangle's center of gravity. It is an important concept because it provides a way to find a balance point of a triangle's mass, assuming uniform density across the triangle's area.

The formula to find the centroid point \( G(x_g, y_g) \) for a triangle with vertices \( (x_1, y_1), (x_2, y_2), \) and \( (x_3, y_3) \) is:
\[ (x_g, y_g) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)\]
This formula shows that the coordinates of the centroid are the averages of the vertices' coordinates. In the case of \( \triangle PQR \), the movement of \( P \) updates the centroid's location, which makes this formula very useful for studying the expression of its locus.

locus equation

Let's explore the concept of a locus, which is a set of all points that satisfy a particular condition. In this exercise, the locus we are interested in is the path traced by the centroid of \( \triangle PQR \) as \( P \) moves along a given line.

To find the locus of the centroid, substitute the coordinates of \( P \) into the centroid formula. Given that \( P \) moves along the line \( 2x - 3y + 4 = 0 \), you express \( y \) in terms of \( x \) to maintain this condition, resulting in \( y = \frac{2}{3}x + \frac{4}{3} \). Substitute this into the expressions for the centroid:

• \( x_g = \frac{x + 4}{3} \)
• \( y_g = \frac{2x + 10}{9} \)

By solving these equations, the locus equation is derived, showing that the traced centroid also follows a linear path.

line equations

Understanding line equations is crucial for analyzing this problem. A line in a plane can be represented by a linear equation, typically written as \( y = mx + c \), where \( m \) is the slope and \( c \) the y-intercept.

In this problem, point \( P \) moves along the line described by \( 2x - 3y + 4 = 0 \). Rearranging gives the equivalent slope-intercept form:
\[ y = \frac{2}{3}x + \frac{4}{3} \]
The slope \( \frac{2}{3} \) implies the line rises slowly compared to its run, which helps shape the trajectory of the centroid's locus.

Finally, the derived centroid locus equation
\[ y_g = \frac{2}{3}x_g + \frac{2}{3} \]
is also in the standard line form, indicating that the locus traces another line with a slope of \( \frac{2}{3} \). With these slopes, we can see that both the path of \( P \) and the locus of the centroid maintain the same directional trend.