Problem 11 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(L\) be the line passing through the point \(P(1,2)\) such that its intercepted segment between the co-ordinate axes is bisected at \(\mathrm{P}\). If \(\mathrm{L}_{1}\) is the line perpendicular to \(\mathrm{L}\) and passing through the point \((-2,1)\), then the point of intersection of \(L\) and \(L_{1}\) is: \(\quad\) [Online April 10,2015\(]\) (a) \(\left(\frac{4}{5}, \frac{12}{5}\right)\) (b) \(\left(\frac{3}{5}, \frac{23}{10}\right)\) (c) \(\left(\frac{11}{20}, \frac{29}{10}\right)\) (d) \(\left(\frac{3}{10}, \frac{17}{5}\right)\)

Step-by-Step Solution

Answer

The intersection point is \(\left(\frac{4}{5}, \frac{12}{5}\right)\).

Step 1: Understand the Problem

We need to find the point of intersection between a line \(L\), which passes through point \(P(1, 2)\) and bisects its intercepted segment between the axes, and another line \(L_1\), which is perpendicular to \(L\) and passes through the point \((-2, 1)\).

Step 2: Determine the Equation of Line L

The line \(L\) intercepts the x-axis and y-axis at points \( (a, 0) \) and \( (0, b) \) respectively. Since point \( (1, 2) \) bisects this segment, it implies \( a/2 = 1 \) and \( b/2 = 2 \). Solving these gives \( a = 2 \) and \( b = 4 \). Thus, the equation of line \(L\) is \( \frac{x}{2} + \frac{y}{4} = 1 \). Simplifying, the equation becomes \(2x + y = 4\).

Step 3: Find the Slope of Line L

Rewriting the equation \(2x + y = 4\) in slope-intercept form, we get \(y = -2x + 4\). Thus, the slope of line \(L\) is \(-2\).

Step 4: Determine the Equation of Line L1

Line \(L_1\) must be perpendicular to \(L\). The perpendicular slope is the negative reciprocal of \(-2\), which is \(\frac{1}{2}\). Using the point \((-2, 1)\), we apply the point-slope form: \(y - 1 = \frac{1}{2}(x + 2)\). Simplifying gives \(y = \frac{1}{2}x + 2\).

Step 5: Solve for the Intersection of L and L1

We solve the equations \( y = -2x + 4 \) and \( y = \frac{1}{2}x + 2 \) simultaneously to find the intersection. Setting \(-2x + 4 = \frac{1}{2}x + 2\), we clear the fractions and find \(-4x + 8 = x + 4\). Solving this gives \(5x = 4\) or \(x = \frac{4}{5}\). Substituting \(x = \frac{4}{5}\) into the equation of \(L\), \(y = -2(\frac{4}{5}) + 4 = \frac{12}{5}\).

Step 6: Verify the Intersection Point

Substitute \(x = \frac{4}{5}\) back into both original equations and confirm \(y = \frac{12}{5}\) for both \(L\) and \(L_1\), confirming that \(\left(\frac{4}{5}, \frac{12}{5}\right)\) is the correct point of intersection.

Key Concepts

Line Equation

The line equation is a fundamental concept in coordinate geometry, and it describes how a straight line can be expressed mathematically. A line in the plane can be defined given certain points or conditions. In the context of the original problem, we first define the line \( L \) by using its intercepted segment between the axes. The points \( (a, 0) \) and \( (0, b) \) represent where the line crosses the x-axis and y-axis, respectively. Since the point \( (1, 2) \) bisects this segment, we have a unique situation where the midpoint is known, allowing us to find the intercepts \( a \) and \( b \). When translated into the standard form, the line has the equation \( 2x + y = 4 \). Understanding how to form line equations based on given conditions is key in solving geometric problems.

Slope-Intercept Form

The slope-intercept form is a very helpful tool in understanding and graphing lines. It is expressed as \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. For line \( L \), the equation \( 2x + y = 4 \) can be rewritten as \( y = -2x + 4 \) to fit this form. Here, \( -2 \) is the slope, which indicates the line descends as you move from left to right, and \( 4 \) is where the line crosses the y-axis. This format makes it easy to identify parallel and perpendicular lines, among other line properties. Knowing how to convert to and from the slope-intercept form is an essential skill in geometry.

Intersection of Lines

The point where two lines meet is known as their intersection. To find the intersection point of two lines, we solve their equations simultaneously. For lines \( L : y = -2x + 4 \) and \( L_1: y = \frac{1}{2}x + 2 \), we equate them to find common \( x \) and \( y \). Solving the equation \(-2x + 4 = \frac{1}{2}x + 2\), we simplify to find \( x = \frac{4}{5} \). By substituting \( x \) back into either line equation, we obtain \( y = \frac{12}{5} \), leading to the intersection point \( \left( \frac{4}{5}, \frac{12}{5} \right) \). This step further illustrates how algebraic manipulation solves geometric problems. Checking this solution against both original line equations ensures the point is correct.

Perpendicular Lines

Perpendicular lines intersect at a right angle (90 degrees), and their slopes are negative reciprocals of each other. In the problem, line \( L_1 \) needs to be perpendicular to \( L \). Line \( L \) has a slope of \(-2\), so the slope of \( L_1 \) must be \( \frac{1}{2} \) because it is the negative reciprocal of \(-2\). Using the point-slope form \( y - y_1 = m(x - x_1) \), we find line \( L_1 \)'s equation as \( y = \frac{1}{2}x + 2 \). Understanding the relationship between perpendicular slopes is crucial in creating and recognizing perpendicular line equations. This concept plays a significant role in fields involving geometry, physics, and engineering.