Problem 70 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If a line \(L\) is perpendicular to the line \(5 x-y=1\), and the area of the triangle formed by the line \(L\) and the coordinate axes is 5, then the distance of line \(L\) from the line \(x+5 y=0\) is: \mathrm{\\{} O n l i n e ~ A p r i l ~ 1 9 , ~ 2 0 1 4 ] ~ (a) \(\frac{7}{\sqrt{5}}\) (b) \(\frac{5}{\sqrt{13}}\) (c) \(\frac{7}{\sqrt{13}}\) (d) \(\frac{5}{\sqrt{7}}\)

Step-by-Step Solution

Answer

The distance is \(\frac{5}{\sqrt{13}}\).

Step 1: Determine the slope of the line perpendicular to L

The equation of the first line is given as \(5x - y = 1\). Rewriting it in slope-intercept form \(y = mx + b\), we have \(y = 5x - 1\). So, the slope \(m_1\) is 5. Since line \(L\) is perpendicular to this line, its slope \(m_2\) should be the negative reciprocal, i.e., \(m_2 = -1/5\).

Step 2: Write the equation of line L

Line \(L\) has a slope of \(-1/5\). Therefore, its equation can be expressed as \(y = -\frac{1}{5}x + c\), where \(c\) is the y-intercept.

Step 3: Calculate x-intercept and y-intercept of line L

The x-intercept is found by setting \(y = 0\) in \(y = -\frac{1}{5}x + c\), giving \(0 = -\frac{1}{5}x + c\) or \(x = 5c\). The y-intercept is \(c\).

Step 4: Set up the area equation for the triangle

The area of the triangle formed by line \(L\), the x-axis, and the y-axis is given as 5. For intercepts \((5c, 0)\) and \((0, c)\), the area of the triangle is \(\frac{1}{2} \times 5c \times c = 5\). This simplifies to \(\frac{5c^2}{2} = 5\).

Step 5: Solve for c

Solving \(\frac{5c^2}{2} = 5\), we multiply both sides by 2, giving \(5c^2 = 10\). Dividing by 5, we get \(c^2 = 2\). Taking the square root, \(c = \sqrt{2}\) or \(c = -\sqrt{2}\).

Step 6: Choose appropriate c value for line L

Since either \(c = \sqrt{2}\) or \(c = -\sqrt{2}\) is acceptable from the area perspective, we proceed with \(c = \sqrt{2}\) to find the distance.

Step 7: Find the perpendicular distance from L to x + 5y = 0

The equation of line \(L\) is \(y = -\frac{1}{5}x + \sqrt{2}\) or rearranged \(x + 5y - 5\sqrt{2} = 0\). The distance between this line and the line \(x + 5y = 0\) is \(\frac{|0 + 5 \sqrt{2}|}{\sqrt{1^2 + 5^2}}\).

Step 8: Calculate the distance

The distance is \(\frac{5\sqrt{2}}{\sqrt{1 + 25}} = \frac{5\sqrt{2}}{\sqrt{26}}\). Simplifying, distance is \(\frac{5}{\sqrt{13}}\).

Key Concepts

Coordinate Geometry

Coordinate geometry, often referred to as analytic geometry, is the study of geometric figures through a coordinate system. This involves representing geometric shapes in a coordinate plane using algebraic equations. A coordinate plane is made up of a horizontal x-axis and a vertical y-axis that divide the plane into four quadrants.
For example, the line equation can be rewritten in the form of coordinates \(y = mx + c\), where \(m\) represents the slope and \(c\) represents the y-intercept. These equations will represent lines on the coordinate plane, allowing us to solve for intersections, distances, and areas.
Understanding coordinate geometry aids in visualizing problems, determining shapes, analyzing geometric properties, and finding the relationships between different figures.

Slope

The concept of slope describes the steepness or inclination of a line. In the coordinate plane, the slope \(m\) of a line is calculated as the ratio of the change in y (vertical change) to the change in x (horizontal change), often expressed as \(m = \frac{\Delta y}{\Delta x}\).
When two lines are perpendicular, the slope of one is the negative reciprocal of the slope of the other. In the exercise, the given line has a slope of 5. Thus, the perpendicular line must have a slope of \(-\frac{1}{5}\).
A solid understanding of slope is crucial as it not only helps in identifying the orientation of lines but also in solving problems related to angles between lines and optimizing distances.

Distance Between Lines

Distance between two parallel lines in a plane can be calculated using their equations. For non-parallel lines, challenges arise, often requiring specific conditions such as perpendicular distances.
In our exercise, we dealt with the distance between line \(L\) and another line. For two lines given in the form \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\), the distance \(d\) is calculated using:\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}}\].
This formula is handy to compute direct distance and is derived from the principles of coordinate geometry and trigonometry. Calculating distances this way helps in applications involving navigation, boundaries, and understanding spatial relationships.

Triangle Area

In coordinate geometry, triangles formed by lines and axes can have their areas determined using intercepts. When a triangle is formed by a line intersecting the x and y-axis at coordinates (x-intercept, 0) and (0, y-intercept), the area \(A\) is given by:\[ A = \frac{1}{2} \times |x-intercept| \times |y-intercept|\].
This formula applies to right-angled triangles with the right angle at the origin. In our case, using intercepts like 5c for x and c for y with the given area of 5 provides a practical equation to solve for intercepts, illustrating properties of geometry in algebraic forms.
Knowing how to calculate the area helps not only in theoretical problems but also in real-world applications involving land, architecture, and design projects.