Determining the Equations of Lines Passing Through a Point and Inclined at 60° to a Given Line or Curve

Introduction

This article explores the methods to determine the equations of lines passing through a specific point and inclined at 60° to a given line or curve. The study involves understanding the slope of a line, the concept of inclination, the equation of a tangent, and the basic principles of calculus. We will consider both linear and non-linear scenarios to provide a comprehensive understanding of the topic.

Given Line and Its Slope

The original problem involves a line given by the equation ( y 1 - sqrt{3}x ). First, let's determine the slope of this line.

Starting with the given line ( y 1 - sqrt{3}x ). The slope ( m ) can be directly extracted from the equation:

[ m -sqrt{3} ]

The slope ( m ) can also be related to the tangent of the angle ( theta ) that the line makes with the positive x-axis.

[ m tan(theta) ]

Given ( m -sqrt{3} ), we know that:

[ tan(theta) -sqrt{3} ]

This implies that:

[ theta 120^circ quad text{or} quad theta 300^circ ]

Lines Inclined at 60° to the Given Line

We now need to find the equations of lines passing through the point (3, -2) and inclined at 60° to the given line.

Scenario 1: Inclination of 60°

If the angle of inclination of the required lines is 60°, we need to calculate the slopes of these lines. The new slope ( m' ) can be found using the angle ( alpha ) between the new line and the given line:

[ m' frac{m tan(60^circ)}{1 - m cdot tan(60^circ)} quad text{or} quad m' frac{m - tan(60^circ)}{1 m cdot tan(60^circ)} ]

Given ( m -sqrt{3} ) and ( tan(60^circ) sqrt{3} ), we get:

[ m' frac{-sqrt{3} sqrt{3}}{1 - (-sqrt{3}) cdot sqrt{3}} frac{0}{1 3} 0 ]

[ m' frac{-sqrt{3} - sqrt{3}}{1 (-sqrt{3}) cdot sqrt{3}} frac{-2sqrt{3}}{1 - 3} frac{-2sqrt{3}}{-2} sqrt{3} ]

Therefore, the equations of the lines passing through (3, -2) with slopes 0 and ( sqrt{3} ) are:

[ y 2 0 quad text{(vertical line)} ]

[ y 2 sqrt{3}(x - 3) Rightarrow y 2 sqrt{3}x - 3sqrt{3} Rightarrow y sqrt{3}x - 3sqrt{3} - 2 ]

Scenario 2: Inclination of 120°

If the angle of inclination is 120°, we need to use the following formula:

[ m' frac{m tan(120^circ)}{1 - m cdot tan(120^circ)} quad text{or} quad m' frac{m - tan(120^circ)}{1 m cdot tan(120^circ)} ]

Given ( m -sqrt{3} ) and ( tan(120^circ) -sqrt{3} ), we get:

[ m' frac{-sqrt{3} (-sqrt{3})}{1 - (-sqrt{3}) cdot (-sqrt{3})} frac{-2sqrt{3}}{1 - 3} frac{-2sqrt{3}}{-2} sqrt{3} ]

[ m' frac{-sqrt{3} - (-sqrt{3})}{1 (-sqrt{3}) cdot (-sqrt{3})} frac{0}{1 3} 0 ]

Therefore, the equations of the lines passing through (3, -2) with slopes 0 and ( sqrt{3} ) are:

[ y 2 0 quad text{(vertical line)} ]

[ y 2 sqrt{3}(x - 3) Rightarrow y 2 sqrt{3}x - 3sqrt{3} Rightarrow y sqrt{3}x - 3sqrt{3} - 2 ]

Lines Passing Through a Point and Tangent to a Curve

If the line ( y 1 - sqrt{3}x ) is meant as a curve, the problem changes. To find lines inclined at 60° to the curve at a specific point, we need to consider the tangent at that point.

For the curve ( y 1 - sqrt{3}x^2 ), we first find the derivative to get the slope of the tangent:

[ frac{dy}{dx} -2sqrt{3}x ]

At the point where ( y 0 ), we have:

[ 0 1 - sqrt{3}x^2 Rightarrow x^2 frac{1}{sqrt{3}} Rightarrow x pm frac{1}{sqrt[4]{3}} ]

If ( x frac{1}{3} ), then:

[ frac{dy}{dx} -2sqrt{3} cdot frac{1}{3} -frac{2sqrt{3}}{3} -frac{2}{sqrt{3}} -frac{2sqrt{3}}{3} ]

The tangent at ( x frac{1}{3} ) is:

[ y - 0 -frac{2sqrt{3}}{3}(x - frac{1}{3}) Rightarrow y -frac{2sqrt{3}}{3}x frac{2sqrt{3}}{9} ]

The angle ( theta ) that this tangent makes with the x-axis is:

[ tan(theta) -frac{2sqrt{3}}{3} Rightarrow theta tan^{-1}(-frac{2sqrt{3}}{3}) approx -56.31^circ ]

The lines inclined at 60° to this tangent will have slopes:

[ m' tan(theta 60^circ) quad text{and} quad m' tan(theta - 60^circ) ]

[ theta 60^circ 0.53996 Rightarrow tan(theta 60^circ) approx 0.0927 Rightarrow y 2 0.0927(x - 3) Rightarrow y 0.0927x - 3cdot 0.0927 - 2 approx 0.0927x - 2.28 ]

[ theta - 60^circ -116.31^circ Rightarrow tan(theta - 60^circ) approx -2.144 Rightarrow y 2 -2.144(x - 3) Rightarrow y -2.144x 6.432 - 2 approx -2.144x 4.432 ]

Conclusion

In conclusion, the equations of lines passing through a specific point and inclined at a given angle to a given line or curve can be determined by understanding the slope of the line and the angle of inclination. The principles of slopes, inclination, and calculus are crucial in solving such problems. Whether the problem involves a linear equation or a curve, the methods remain fundamentally similar but may involve more complex calculations.

For any further questions or deeper understanding, refer to the detailed calculations and principles described in this article. If you have any specific queries or need further assistance, feel free to reach out!