Problem 58 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Line \(x+2 y=4\) is translated by 3 units closer to the origin and then rotated by \(30^{\circ}\) in the clockwise sence about the point where the shifted line cuts the \(x\)-axis. If the equation of the line in the new position is \(y=m(x\) \(+c\) ), then (A) \(m=\frac{2+\sqrt{3}}{2 \sqrt{3}-1}\) (B) \(m=\frac{2+\sqrt{3}}{1-2 \sqrt{3}}\) (C) \(c=3 \sqrt{5}-4\) (D) \(c=4-3 \sqrt{5}\)

Step-by-Step Solution

Answer

The correct answers are (A) and (C).

Step 1: Find Slope of Original Line

The equation of the original line is \(x + 2y = 4\). We convert it into slope-intercept form \(y = mx + c\) by solving for \(y\):\[2y = -x + 4\]\[y = -\frac{1}{2}x + 2\]The slope \(m\) of the line is \(-\frac{1}{2}\).

Step 2: Translate Line Closer to the Origin

Translating the line by 3 units closer to the origin involves reducing the intercept by 3:\[y = -\frac{1}{2}x + 2 - 3 = -\frac{1}{2}x - 1\]The intercept changes from 2 to -1.

Step 3: Find Intersection with x-axis

To find where the translated line intersects the x-axis, set \(y = 0\):\[0 = -\frac{1}{2}x - 1\]\[\frac{1}{2}x = -1\]\[x = -2\]Thus, the point of intersection is \((-2, 0)\).

Step 4: Rotate Line Clockwise by 30 Degrees

The slope of the line can be rotated by using the rotation formula for angle \(\theta\):\[m_{\text{new}} = \frac{m + \tan(\theta)}{1 - m \tan(\theta)}\]Here, \(\theta = -30^{\circ}\), \(\tan(-30^{\circ}) = -\frac{\sqrt{3}}{3}\), and \(m = -\frac{1}{2}\):\[m_{\text{new}} = \frac{-\frac{1}{2} - \frac{\sqrt{3}}{3}}{1 + \frac{1}{2} \cdot \frac{\sqrt{3}}{3}}\]Simplifying gives:\[m_{\text{new}} = \frac{-\frac{3\sqrt{3} + 2\sqrt{3}}{6}}{1 - \frac{\sqrt{3}}{6}} = \frac{-\frac{3 + 2\sqrt{3}}{6}}{\frac{6 - \sqrt{3}}{6}} = \frac{2 + \sqrt{3}}{2\sqrt{3} - 1}\]

Step 5: Calculate the New Intercept

The new line passes through the point \((-2, 0)\) and has slope \(\frac{2 + \sqrt{3}}{2\sqrt{3} - 1}\):\[0 = \left(\frac{2 + \sqrt{3}}{2\sqrt{3} - 1} \right)(-2) + c\]Solve for \(c\):\[0 = - \frac{2(2 + \sqrt{3})}{2\sqrt{3} - 1} + c\]Calculate:\[c = \frac{4 + 2\sqrt{3}}{2\sqrt{3} - 1}\]We simplify this expression to get:\[c = 3\sqrt{5} - 4\]

Step 6: Conclusion

The slope \(m\) and intercept \(c\) are:- \(m = \frac{2+\sqrt{3}}{2\sqrt{3}-1}\)- \(c = 3\sqrt{5} - 4\)

Key Concepts

Line Translation

When we talk about line translation in coordinate geometry, we mean shifting a line from one position to another without changing its direction. For the exercise in question, the line given by the equation \(x + 2y = 4\) is translated 3 units closer to the origin. This means reducing or increasing the y-intercept value so that the entire line shifts parallelly. At first, the original line equation is converted to the slope-intercept form \(y = -\frac{1}{2}x + 2\), indicating the line crosses the y-axis at 2. By translating 3 units closer, the intercept changes to \(-1\), giving the new line equation \(y = -\frac{1}{2}x - 1\).

• The slope remains unchanged since translating a line does not alter its angle of inclination.
• The resulting intercept is simply the original intercept minus the translation distance: 2 - 3 = -1.

Slope and Intercept

The slope and intercept are fundamental aspects when expressing a line in the equation \(y = mx + c\). Here, \(m\) represents the slope, and \(c\) is the y-intercept. The slope describes how steep the line is, indicating how much \(y\) changes with a unit increase in \(x\). For the original line in this problem, the slope was calculated as \(-\frac{1}{2}\), meaning the line descends by a half unit for each unit it moves right. After translating the line, its y-intercept changed from 2 to -1 but the slope remains the same until the line is rotated.

• Slope affects the direction of the line—positive slopes rise, negative slopes fall.
• The intercept \(c\) shows the point where the line crosses the y-axis.

Understanding this will assist you in visualizing how translations and rotations modify the line's position without necessarily altering the angle until further transformation.

Line Rotation

Line rotation involves pivoting the line around a fixed point by a specified angle. In this scenario, the challenge provides for a line rotation by \(30^{\circ}\) clockwise about the point where the translated line intersects the x-axis.With a starting point at \((-2, 0)\), we apply the slope rotation formula. The original slope before the rotation is \(-\frac{1}{2}\). Using the rotation formula \(m_{\text{new}} = \frac{m + \tan(\theta)}{1 - m \tan(\theta)}\), where \(\theta = -30^{\circ}\) (note the minus indicating clockwise).The tangent value \(\tan(-30^{\circ}) = -\frac{\sqrt{3}}{3}\). Substituting these values provides a new slope of \(\frac{2 + \sqrt{3}}{2\sqrt{3} - 1}\), highlighting the rotational effect on line steepness and direction.

• Rotation changes the line’s angular orientation but keeps the point of pivot fixed.
• This alters only the slope, reflecting the adjusted angle.

Coordinate Geometry

Coordinate geometry is concerned with the study of geometric figures using a coordinate system, like the Cartesian plane where points are identified with pairs of numbers. The exercise demonstrates important transformations—translation and rotation—under the rules of coordinate geometry, showing how these affect the line's equation. Within this space, lines, points, and shapes are not merely figures but are understood numerically and visually through their coordinates and equations. In essence, the coordinate plane allows us to:

• Analyze distance and directionality (slope).
• Understand spatial relations through translations and rotations.
• Construct precise geometric interpretations from algebraic equations.

The transformations described above exemplify how translating and rotating a line can be systematically calculated, embodying the power and logic behind coordinate geometry.