Problem 42 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The coordinates of the point which is at unit distance from the lines \(L_{1} \equiv 3 x-4 y+1=0\) and \(L_{2} \equiv 8 x+6 y+\) \(1=0\) and lies below \(L_{1}\) and above \(L_{2}\) are (A) \(\left(\frac{6}{5}, \frac{1}{10}\right)\) (B) \(\left(\frac{6}{5},-\frac{1}{10}\right)\) (C) \(\left(\frac{6}{5}, \frac{1}{5}\right)\) (D) \(\left(\frac{6}{5},-\frac{1}{5}\right)\)

Step-by-Step Solution

Answer

Option (A): \(\left(\frac{6}{5}, \frac{1}{10}\right)\) solves the conditions.

Step 1: Understanding the Problem

We need to find a point which is at a unit distance from both lines, lies below line \(L_1\), and above line \(L_2\). The distance must be calculated perpendicular to the lines.

Step 2: Calculate Distance from Line 1

The distance from a point \((x, y)\) to a line \(Ax + By + C = 0\) is given by the formula \(\frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}\). For line \(L_1: 3x - 4y + 1 = 0\), the perpendicular distance formula becomes \(\frac{|3x - 4y + 1|}{\sqrt{3^2 + (-4)^2}} = 1\). Simplifying, we get \(\frac{|3x - 4y + 1|}{5} = 1\), leading to two equations: \(|3x - 4y + 1| = 5\).

Key Concepts

Perpendicular Distance Formula

When studying coordinate geometry, it's important to understand the concept of finding the shortest distance from a point to a line. This is known as the perpendicular distance. We use the perpendicular distance formula to calculate this.
The formula to find the perpendicular distance from a point \((x_1, y_1)\) to a line given by the equation \(Ax + By + C = 0\) is:

• \(\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)

This formula measures how far off a point is from lying on the line, following a perpendicular path. In simpler terms, it’s like measuring how far you are from a wall by making a direct perpendicular line to it.
For any point that needs to be a specific distance from the line, you'd set the formula equal to that distance and solve for the point's coordinates.
In our textbook problem, you're asked to find a point equidistant (unit distance) from two lines. This setup requires using the perpendicular distance formula for each line. You then solve these equations simultaneously to find the correct coordinates that satisfy both conditions.

Equation of a Line

Understanding how to interpret and utilize the equation of a line is key in coordinate geometry. The general form of a line's equation is \(Ax + By + C = 0\). Components such as

• \(A\), the coefficient of \(x\)
• \(B\), the coefficient of \(y\)
• \(C\), the constant term

play a critical role in defining the orientation of the line in the plane.
In different contexts, these coefficients help you determine the slope and position of the line:

• The slope \(m\) is calculated as \(-\frac{A}{B}\). It signifies how steep the line is.
• Shifting \(C\) changes the line's position without altering its slope.

By examining the lines \(L_1: 3x - 4y + 1 = 0\) and \(L_2: 8x + 6y + 1 = 0\), you assess their slopes and how they relate spatially to the point sought.
Understanding these aspects enables the assessment of whether a point lies above, below, or near a line based on its positioning dictated by its equation. This foundational knowledge is crucial for solving problems that involve finding distances or intersections of lines in geometry.

Unit Distance

The term *unit distance* refers to a distance of one unit length in whatever scale or measurement you’re using. In mathematics, especially coordinate geometry, a unit distance allows you to standardize comparisons and measurements.
When a point is at a unit distance from a line, it means that if you were to "walk" directly perpendicular to the line for one unit, you would reach the point.
This is a common measurement when solving problems involving distances in geometric spaces — it simplifies application of the distance formulas and helps in achieving calculations in a more straightforward manner.
In the given textbook problem, the challenge is to find a point at this specific unit distance from two given lines. This involves using the distance formulas and confirming that the solution satisfies both linear conditions of staying within one unit's span from each line.