Problem 23 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A point on the straight line, \(3 x+5 y=15\) which is equidistant from the coordinate axes will lie only in : [April 8, 2019 (I)] (a) \(4^{\text {th }}\) quadrant (b) \(1^{\text {st }}\) quadrant (c) \(1^{\text {st }}\) and \(2^{\text {wd }}\) quadrants (d) \(1^{\text {st }}, 2^{\text {nd }}\) and \(4^{\text {th }}\) quadrants

Step-by-Step Solution

Answer

The point lies only in the 1st quadrant, so the answer is (b).

Step 1: Understand the Condition

The point is equidistant from the coordinate axes, which means the absolute value of its x-coordinate is equal to the absolute value of its y-coordinate. Therefore, the point has coordinates (a, a) or (-a, -a).

Step 2: Substitute into Line Equation

Substitute (a, a) into the line equation: \(3a + 5a = 15\), which simplifies to \(8a = 15\). Substitute (-a, -a): \(3(-a) + 5(-a) = 15\), simplifies to \(-8a = 15\).

Step 3: Solve the Equations

From \(8a = 15\), we find \(a = \frac{15}{8}\). Evaluating \(-8a = 15\), gives \(a = -\frac{15}{8}\). Thus, the points are \(\left(\frac{15}{8}, \frac{15}{8}\right)\) and \(\left(-\frac{15}{8}, -\frac{15}{8}\right)\).

Step 4: Determine Quadrants

The point \(\left(\frac{15}{8}, \frac{15}{8}\right)\) is in the 1st quadrant because both coordinates are positive. The point \(\left(-\frac{15}{8}, -\frac{15}{8}\right)\) is in the 3rd quadrant because both coordinates are negative.

Key Concepts

Quadrants

In coordinate geometry, the plane is divided into four regions known as quadrants. These quadrants are labeled in a counter-clockwise direction starting from the top right:

• The 1st quadrant, where both the x and y coordinates are positive.
• The 2nd quadrant, where x is negative and y is positive.
• The 3rd quadrant, where both the x and y coordinates are negative.
• The 4th quadrant, where x is positive and y is negative.

Understanding the concept of quadrants helps in determining the location of points on a coordinate plane. For example, the point \( rac{15}{8}, rac{15}{8} \) is found in the 1st quadrant because both coordinates are positive. This knowledge is integral for solving many problems in coordinate geometry, including the one where a point is equidistant from the coordinate axes.

Line Equation

The equation of a line is a mathematical way to represent all the points that lie on a straight line in a plane. A typical line equation is written in the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. In the given exercise, the line equation is \( 3x + 5y = 15 \).
To find specific points on this line, we can substitute values for either \( x \) or \( y \) and solve for the other variable. For instance, if we let the point on the line be equidistant from the x and y axes, the point will have coordinates like \( (a, a) \) or \( (-a, -a) \). After substituting these into the line equation, we can solve for the specific values of \( a \) that place the points precisely on the line. This step-by-step substitution allows us to find the intersection of points and understand the behavior and position of the line in the coordinate system.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, combines algebra and geometry to examine geometric figures like lines, points, and curves in a numerical way. It uses the coordinate plane to provide a bridge between algebraic equations and geometric shapes.
In coordinate geometry, a point is identified by an ordered pair \( (x, y) \) on the coordinate plane. Each of these points can be used to satisfy a specific condition or equation, such as being equidistant from the coordinate axes as in the exercise discussed. The key here is to recognize how algebraic manipulation of these coordinates can determine crucial properties such as distance and position.
This area of mathematics is essential for solving various practical and theoretical problems, as it allows for the visualization and calculation of the properties of geometric figures with precision, harnessing the power of algebra.