Problem 21 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\mathrm{C}\) be the centroid of the triangle with vertices \((3,-1)\), \((1,3)\) and \((2,4) .\) Let \(\mathrm{P}\) be the point of intersection of the lines \(x+3 y-1=0\) and \(3 x-y+1=0 .\) Then the line passing through the points \(\mathrm{C}\) and \(\mathrm{P}\) also passes through the point: [Jan. \(9,2020(\mathrm{I})]\) (a) \((-9,-6)\) (b) \((9,7)\) (c) \((7,6)\) (d) \((-9,-7)\)

Step-by-Step Solution

Answer

(a) (-9, -6)

Step 1: Find the Centroid C

The coordinates of the centroid \(C\) of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) are given by: \(\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)\). For the vertices \((3, -1), (1, 3), (2, 4)\), calculate:\[C = \left( \frac{3+1+2}{3}, \frac{-1+3+4}{3} \right) = \left( \frac{6}{3}, \frac{6}{3} \right) = (2, 2)\]

Step 2: Find Point P

Solve the system of equations given by the lines \(x + 3y - 1 = 0\) and \(3x - y + 1 = 0\). Start by solving for \(x\) and \(y\):From the first equation, express \(x\) in terms of \(y\):\[ x = 1 - 3y \]Substitute this expression into the second equation:\[ 3(1 - 3y) - y + 1 = 0 \]\[ 3 - 9y - y + 1 = 0 \]\[ -10y = -4 \]\[ y = \frac{2}{5} \]Substitute \(y = \frac{2}{5}\) back into \(x = 1 - 3y\):\[ x = 1 - 3\left( \frac{2}{5} \right) = 1 - \frac{6}{5} = \frac{-1}{5} \]So \(P = \left( \frac{-1}{5}, \frac{2}{5} \right) \).

Step 3: Determine the Equation of Line CP

Use the point-slope form to find the line equation passing through \(C(2,2)\) and \(P\left(\frac{-1}{5}, \frac{2}{5}\right)\). The slope \(m\) is given by:\[ m = \frac{2 - \frac{2}{5}}{2 - \frac{-1}{5}} = \frac{\frac{10}{5} - \frac{2}{5}}{\frac{10}{5} + \frac{1}{5}} = \frac{\frac{8}{5}}{\frac{11}{5}} = \frac{8}{11} \]Using the point-slope form \( y - y_1 = m(x - x_1) \) with \((x_1, y_1) = (2, 2)\), we have:\[ y - 2 = \frac{8}{11}(x - 2) \] which simplifies to:\[ 11(y - 2) = 8(x - 2) \]\[ 11y - 22 = 8x - 16 \]\[ 8x - 11y + 6 = 0 \]

Step 4: Verify which given point lies on Line CP

Substitute each of the given points into the line equation \(8x - 11y + 6 = 0\) to see which one satisfies it:For option (a) \((-9, -6)\):\[ 8(-9) - 11(-6) + 6 = -72 + 66 + 6 = 0 \]The equation holds true. Thus, \((-9, -6)\) lies on the line.

Key Concepts

Point of Intersection

In the context of geometry, the "Point of Intersection" refers to the point where two lines meet each other. To find this point, you need to solve the system of equations that represent the lines. This involves finding the values of the variables (typically \(x\) and \(y\)) that satisfy both equations simultaneously.
For example, in our problem, we have two lines\( x + 3y - 1 = 0 \) and \( 3x - y + 1 = 0 \). When these equations are solved as a system, they reveal the coordinates \( (x, y) \) of their intersection point, \( P \).
Follow these simple steps to find a point of intersection:

• Express one variable in terms of the other from one of the equations.
• Substitute this expression into the second equation.
• Solve the resulting equation for the second variable.
• Substitute back to find the first variable.

This results in the coordinates of the intersection point, making it an essential concept for analyzing where lines cross each other.

Slope of a Line

The "Slope of a Line" is a measure that describes the steepness and the direction of a line. Calculating the slope is crucial when you want to write an equation of a line through two points or analyze linear relationships.
In mathematical terms, the slope \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \). This formula tells us how much the \(y\)-coordinate changes for a unit change in the \(x\)-coordinate.
For instance, in calculating the slope of the line passing through \(C(2,2)\) and \(P\left(\frac{-1}{5}, \frac{2}{5}\right)\), we compute:

• Find the difference in the \(y\)-coordinates: \(2 - \frac{2}{5}\)
• Find the difference in the \(x\)-coordinates: \(2 - \frac{-1}{5}\)
• Calculate \( m = \frac{\frac{8}{5}}{\frac{11}{5}} = \frac{8}{11} \).

The slope \(\frac{8}{11}\) informs us about the line's incline, pointing upwards as it moves from left to right.

Equation of a Line

Understanding how to derive the "Equation of a Line" is foundational in geometry and algebra. When given a point and a slope, or two points, you can determine a line's equation, describing all the points along the line.
The point-slope form of the equation of a line is: \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope. This form is particularly helpful when you already know these values.
In our exercise, by using a centroid point \((2, 2)\) and a point of intersection \(\left(\frac{-1}{5}, \frac{2}{5}\right)\), we found the equation of the line, after establishing the slope: \( y - 2 = \frac{8}{11}(x - 2) \).
This equation can be further simplified or transformed into other forms such as the slope-intercept form \(y = mx + b\) or standard form \(Ax + By + C = 0\), depending on the needs of the problem.

System of Linear Equations

A "System of Linear Equations" consists of two or more linear equations involving the same set of variables. Solving this system means finding the values of these variables that satisfy all equations at once.
This is a powerful tool used to find the specific points where certain conditions are true, such as the point of intersection between lines.
To solve a system of linear equations, follow these steps:

• Use substitution or elimination methods to reduce the number of variables within the equations.
• For substitution, isolate one variable in one equation and substitute into the other equation.
• For elimination, add or subtract equations to eliminate one variable, making it easier to solve for the other.

In our scenario, we applied substitution by expressing \(x\) from one equation and then substituting it into the other to find \(y\). Solving this system gave us the coordinates of the intersection point \((\frac{-1}{5}, \frac{2}{5})\). This method illustrates the capability of systems of equations in resolving points of convergence between algebraic expressions.