Problem 55 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the line, \(2 x-y+3=0\) is at a distance \(\frac{1}{\sqrt{5}}\) and \(\frac{2}{\sqrt{5}}\) from the lines \(4 x-2 y+\alpha=0\) and \(6 x-3 y+\beta=0\), respectively, then the sum of all possible value of \(\alpha\) and \(B\) is

Step-by-Step Solution

Answer

The sum of all possible values is 24.

Step 1: Understand the Problem

We need to find the sum of all possible values of \(\alpha\) and \(\beta\) such that the line \(2x - y + 3 = 0\) is at given distances \(\frac{1}{\sqrt{5}}\) and \(\frac{2}{\sqrt{5}}\) from the lines \(4x - 2y + \alpha = 0\) and \(6x - 3y + \beta = 0\) respectively.

Step 2: Review the Distance Formula

The perpendicular distance \(d\) from a line \(ax + by + c = 0\) to a line \(ax + by + c' = 0\) is given by the formula: \[d = \frac{|c-c'|}{\sqrt{a^2 + b^2}}\] where \(a, b\) represent the coefficients of \(x\) and \(y\).

Step 3: Calculate Distance from First Line

For the lines \(2x - y + 3 = 0\) and \(4x - 2y + \alpha = 0\), using \(d = \frac{1}{\sqrt{5}}\), we apply the distance formula to find \(\alpha\):\[\frac{|3-\alpha|}{\sqrt{4^2 + (-2)^2}} = \frac{1}{\sqrt{5}} \Rightarrow \frac{|3-\alpha|}{2\sqrt{5}} = \frac{1}{\sqrt{5}}\]Simplifying, we get \(|3-\alpha| = 2\Rightarrow \alpha = 5 \text{ or } \alpha = 1\).

Step 4: Calculate Distance from Second Line

For the lines \(2x - y + 3 = 0\) and \(6x - 3y + \beta = 0\), using \(d = \frac{2}{\sqrt{5}}\), apply the distance formula to find \(\beta\):\[\frac{|3-\beta|}{\sqrt{6^2 + (-3)^2}} = \frac{2}{\sqrt{5}} \Rightarrow \frac{|3-\beta|}{3\sqrt{5}} = \frac{2}{\sqrt{5}}\]Simplifying, we get \(|3-\beta| = 6\Rightarrow \beta = 9 \text{ or } \beta = -3\).

Step 5: Compute the Sum of All Possible Values

Now, we need to find the sum of all possible combinations of \(\alpha\) and \(\beta\). This gives us four pairs: \((5,9), (5,-3), (1,9), (1,-3)\). The sums for these pairs are: \(5+9=14\), \(5-3=2\), \(1+9=10\), \(1-3=-2\). Thus, the sum of all possible sums is \(14+2+10-2 = 24\).

Key Concepts

Perpendicular distance formula

The perpendicular distance formula is essential in determining the shortest distance between two parallel lines. It's a direct measure from one line to another without deviating. To calculate this distance between two lines given in the form \( ax + by + c = 0 \) and \( ax + by + c' = 0 \), we use:

• \( d = \frac{|c - c'|}{\sqrt{a^2 + b^2}} \)

Here, \( |c - c'| \) represents the absolute difference in the constant terms of the equations. The denominator, \( \sqrt{a^2 + b^2} \), is simply the magnitude of the normal vector (coefficients of \(x\) and \(y\)) of the line. This formula is useful because it simplifies calculating the exact distance without needing perpendicular intersection points. Remember, parallel lines have the same slope, which makes this formula applicable.

Parallel lines equations

In geometry, parallel lines are lines in a plane that never intersect or meet. They have identical slopes, meaning their inclination relative to the x and y-axis is the same. If you're given the equations of two lines, confirming their parallelism involves checking:

• The slopes (ratios of the coefficients of \(x\) and \(y\)) are equal.

For example, lines with equations \( y = mx + c \) and \( y = mx + c' \) are parallel because they both have slope \( m \). In addition, lines like \( 4x - 2y + \alpha = 0 \) and \( 6x - 3y + \beta = 0 \) can be verified for parallelism by simplifying their ratios of coefficients \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\). Knowing this helps us apply the perpendicular distance formula to solve problems related to parallel lines.

Mathematical problem-solving

Mathematical problem-solving involves identifying known parameters, applying relevant formulas, and logically deducing the required values. In this context, it's crucial to:

• Understand the problem: Identify given, required, and known entities.
• Identify applicable formulas: Like the perpendicular distance formula for parallel lines.
• Solve step-by-step: Use algebraic manipulation and computations sequentially to find the solution.

This exercise requires applying the perpendicular distance formula, making calculations to determine constants \(\alpha\) and \(\beta\), and then summing various possibilities. A methodical approach ensures accuracy and comprehensive understanding of solving parallel line distance problems. This strategic mindset is valuable not just in geometry but across all mathematical analyses.