Question

If \(2 \mathrm{x}+2 \mathrm{y}-5=0\) is the equation of the line containing one of the sides of an equilateral triangle and \((1,2)\) is one vertex, then find the equations of the lines containing the other two sides. (a) \(\mathrm{y}=(2+\sqrt{3}) \mathrm{x}-\sqrt{3}, \mathrm{y}=(2+\sqrt{3}) \mathrm{x}+\sqrt{3}\) (b) \(y=(2-\sqrt{3}) x-\sqrt{3}, y=(2+\sqrt{3}) x+\sqrt{3}\) (c) \(y=(2-\sqrt{3}) x+\sqrt{3}, y=(2+\sqrt{3}) x-\sqrt{3}\) (d) \(y=(2+\sqrt{3}) x+\sqrt{3}, y=(2+\sqrt{3}) x-\sqrt{3}\)

Short answer

The short answer is: (c) \(y = (2 - \sqrt{3})x + \sqrt{3}\) and \(y = (2 + \sqrt{3})x - \sqrt{3}\).

Step-by-step solution

Find the slope of the given equation

First, rewrite the given equation in the form y = mx + b, where m is the slope and b is the y-intercept. The given equation is: \(2x + 2y - 5 = 0\) Rewrite it to find y: \(y = -x + \frac{5}{2}\) Thus, the slope of the given equation is -1.

Find the slopes of the other two sides

To find the slope of the other two sides, we need to consider a 60-degree angle with the given side. The tangent of the angle between two lines with slopes m1 and m2 is given by the formula: \(\tan{\theta} = \frac{m1 - m2}{1 + m1 \cdot m2}\) For a 60-degree angle, \(\tan(60) = \sqrt{3}\). Substitute \(m1 = -1\) (slope of the given side) into the formula: \(\sqrt{3} = \frac{-1 - m2}{1 - m2}\) Solve for m2, which gives two possible slopes: \(m2 = 2 - \sqrt{3} \text{ or } m2 = 2 + \sqrt{3}\)

Find the equations of the other two sides

Given the slopes m2 and the given vertex (1, 2), we can find the point-slope form of the lines containing the other two sides: For m2 = \(2 - \sqrt{3}\) \(y - 2 = (2 - \sqrt{3})(x - 1)\) Which simplifies to: \(y = (2 - \sqrt{3})x + \sqrt{3}\) And for m2 = \(2 + \sqrt{3}\) \(y - 2 = (2 + \sqrt{3})(x - 1)\) Which simplifies to: \(y = (2 + \sqrt{3})x - \sqrt{3}\) So, the equations of the lines containing the other two sides are: \(y = (2 - \sqrt{3})x + \sqrt{3}\) and \(y = (2 + \sqrt{3})x - \sqrt{3}\) And the answer is (c).

Key concepts

Understanding the Slope of a Line

The slope of a line is a measure of its steepness. It is usually represented by the letter "m" in the equation of a line in slope-intercept form:

• For example, in the line equation \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
• Visually, the slope describes how far a line will rise vertically when it runs horizontally.

To determine the slope from an equation, we rearrange it into the slope-intercept form. In our exercise, the given line equation is \( 2x + 2y - 5 = 0 \).
By rearranging it to \( y = -x + \frac{5}{2} \), we find that the slope \( m = -1 \).
This means for every unit the line moves to the right, it moves one unit downwards.
This insight will help us find slopes of other lines perpendicular or at a specific angle.

Formulating the Equation of a Line

Once we know the slope of a line, the next step is to find its equation. This is particularly easy when we have a known point on the line.

• We use the point-slope form of a line equation: \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a point on the line.
• From this, we can rearrange to get the line equation in y = mx + b form.

In the practice problem, we're given a vertex of the triangle at the point \( (1, 2) \).
Using the slopes we derived—\( 2 - \sqrt{3} \) and \( 2 + \sqrt{3} \)—we substitute into the point-slope form.
For the slope \( 2 - \sqrt{3} \), the equation becomes \( y - 2 = (2 - \sqrt{3})(x - 1) \).
This simplifies to \( y = (2 - \sqrt{3})x + \sqrt{3} \).
Demonstrating how changes in slope affect line orientation and position.

Calculating the Angle Between Lines

Understanding the angle between lines is crucial for many geometric applications, including finding side lengths in shapes like triangles.

• The angle between two lines can be determined using their slopes \( m_1 \) and \( m_2 \).
• The formula \( \tan(\theta) = \frac{m_1 - m_2}{1 + m_1 \cdot m_2} \) is used, where \( \theta \) is the angle between them.

In our problem, the lines form a 60-degree angle with each other. Knowing \( \tan(60) = \sqrt{3} \), we set up the equation with the known slope of \( -1 \) (given line's slope) to find the other two slopes.
Simplifying gives us slopes of \( 2 - \sqrt{3} \) and \( 2 + \sqrt{3} \), which account for the equilateral triangle's equal angles.
Understanding this process aids in solving complex geometric problems by primarily using slope relationships.