Problem 58 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Lines are drawn parallel to the line \(4 x-3 y+2=0\), at a distance \(\frac{3}{5}\) from the origin. Then which one of the following points lies on any of these lines ? [April 10, 2019 (II)] (a) \(\left(-\frac{1}{4}, \frac{2}{3}\right)\) (b) \(\left(\frac{1}{4},-\frac{1}{3}\right)\) (c) \(\left(\frac{1}{4}, \frac{1}{3}\right)\) (d) \(\left(-\frac{1}{4},-\frac{2}{3}\right)\)

Step-by-Step Solution

Answer

Option (b) \(\left(\frac{1}{4}, -\frac{1}{3}\right)\) is on the line \(4x - 3y - 1 = 0\).

Step 1: Find the slope of the given line

The given line equation is in the form \(4x - 3y + 2 = 0\). Convert it to the slope-intercept form \(y = mx + c\).\[4x - 3y + 2 = 0 \implies 3y = 4x + 2 \implies y = \frac{4}{3}x + \frac{2}{3}\].The slope \(m\) of the line is \(\frac{4}{3}\).

Step 2: Calculate perpendicular distance from the origin

The perpendicular distance \(d\) from the line \(Ax + By + C = 0\) to the origin \((0,0)\) is given by \(\frac{|C|}{\sqrt{A^2 + B^2}}\).Substitute \(A = 4\), \(B = -3\), and \(C = 2\):\[d = \frac{|2|}{\sqrt{4^2 + (-3)^2}} = \frac{2}{\sqrt{16 + 9}} = \frac{2}{5}\].

Step 3: Set up equations for parallel lines

Lines parallel to the given line will have the same slope \(\frac{4}{3}\). If a line is distance \(\frac{3}{5}\) from the origin, and the original line is distance \(\frac{2}{5}\), the required distance \(k\) from the line to the origin will be given by \(|k - \frac{2}{5}| = \frac{3}{5}\).

Step 4: Solve for possible values of 'C'

Solving \(|k - \frac{2}{5}| = \frac{3}{5}\) yields two cases: 1. \(k - \frac{2}{5} = \frac{3}{5}\): \(k = 1\)2. \(k - \frac{2}{5} = -\frac{3}{5}\): \(k = -\frac{1}{5}\)Thus, the lines are: \(4x - 3y + 1 = 0\) and \(4x - 3y - 1 = 0\).

Step 5: Verify options with the new line equations

Check which point satisfies either line equation \(4x - 3y + 1 = 0\) or \(4x - 3y - 1 = 0\). Let's try each option:- Option (a): Substituting into both lines does not satisfy.- Option (b): \(4(\frac{1}{4}) - 3(-\frac{1}{3}) - 1 = 0\) satisfies \(4x - 3y - 1 = 0\).- Option (c) and (d): Substituting does not satisfy either line.

Key Concepts

Parallel Lines

Parallel lines are lines in a plane that never intersect, no matter how far they are extended. This happens because they always have the same slope. In analytical geometry, if you have a line given in the equation form of \(Ax + By + C = 0\), any other line parallel to it will have the same coefficients \(A\) and \(B\), while \(C\) is the only variable that changes.

• If you convert the equation to the slope-intercept form, \(y = mx + c\), the parallel line shares the same slope \(m\).
• The distance between parallel lines is determined by the difference in their \(C\) values divided by the norm of the vector \((A, B)\).

In practical problems, creating parallel lines often involves calculating positions, ensuring the same slope is maintained, which means the elements that change involve perpendicular offsets, not direction.

Perpendicular Distance

The perpendicular distance from a point to a line is the shortest distance between the line and that point. This is particularly useful when you know the line equation and need to gauge proximity from a specific location, like the origin \((0,0)\).
For a line \(Ax + By + C = 0\), the perpendicular distance \(d\) from the origin is calculated using: \[d = \frac{|C|}{\sqrt{A^2 + B^2}}\].

• This formula stems from the algebraic manipulation of point-line distance and refreshingly circumvents more complex coordinate positioning.
• The value within the absolute brackets \(|C|\) ensures that distance is positive and treats distance as merely a scalar measure.

Whether for geometry applications or ensuring accuracy in analytical problems, knowing how to apply perpendicular distance is indispensable.

Equation of a Line

An equation of a line in a plane can describe the line's path in space, often in formats like the general form \(Ax + By + C = 0\) or the slope-intercept form \(y = mx + c\).
Converting between these forms is straightforward and helps in analyzing line equations:

• Start by isolating \(y\) in the general equation \(Ax + By + C = 0\), leading you to the form \(y = \frac{-A}{B}x - \frac{C}{B}\).
• By recognizing this format, the slope \(m\) becomes \(-\frac{A}{B}\), and the intercept \(c\) transforms into \(-\frac{C}{B}\).

Recognizing these forms allows for straightforward calculations like finding perpendicular distances, setting up parallel lines, and determining whether specific points fall on a line.

Slope-Intercept Form

The slope-intercept form, \(y = mx + c\), is a popular equation representing a straight line, making it easy to determine both the slope \(m\) and the y-intercept \(c\).
Using this form:

• The slope \(m\) reveals the steepness or angle of a line relative to the horizontal plane.
• The y-intercept \(c\) indicates where the line crosses the y-axis, serving as a pivotal point in drafting the graph.

This form simplifies understanding and graphing, providing immediate insights into how a line behaves across a coordinate space. It's especially handy for transforming equations for easier manipulation and offers clear insights, particularly when analyzing and crafting parallel and perpendicular lines.