Understanding the Equilateral Triangle Geometry Problem

This problem involves finding the equation of a line that meets specific criteria related to an equilateral triangle ABC. We are given the side length, the position and orientation of one side (AB) on the y-axis, and the location of the base's midpoint. We then need to find the equation of a line passing through a specific point and parallel to another side (AC) of the triangle.

Determining the Coordinates of the Vertices

The base AB of the equilateral triangle has a side length of 8 cm. It lies along the y-axis, and its midpoint is at the origin (0, 0). Since B lies above the origin and the total length of AB is 8 cm, the coordinates of A and B can be determined:

• The midpoint of AB is (0, 0).
• The length of AB is 8 cm.
• Since the midpoint is at the origin, A and B must be 4 cm away from the origin along the y-axis.
• B is above the origin, so its y-coordinate is positive. Thus, B is at (0, 4).
• A is below the origin, so its y-coordinate is negative. Thus, A is at (0, -4).

Now we need to find the coordinates of vertex C. Since ABC is an equilateral triangle, the distance from C to A must be 8 cm, and the distance from C to B must also be 8 cm. Let the coordinates of C be \((x_C, y_C)\).

Using the distance formula:

Distance AC \(=\) 8

\(\sqrt{(x_C - 0)^2 + (y_C - (-4))^2} = 8\)

\(\sqrt{x_C^2 + (y_C + 4)^2} = 8\)

\(x_C^2 + (y_C + 4)^2 = 64 \quad \text{(Equation 1)}\)

Distance BC \(=\) 8

\(\sqrt{(x_C - 0)^2 + (y_C - 4)^2} = 8\)

\(\sqrt{x_C^2 + (y_C - 4)^2} = 8\)

\(x_C^2 + (y_C - 4)^2 = 64 \quad \text{(Equation 2)}\)

Equating Equation 1 and Equation 2:

\(x_C^2 + (y_C + 4)^2 = x_C^2 + (y_C - 4)^2\)

\((y_C + 4)^2 = (y_C - 4)^2\)

\(y_C^2 + 8y_C + 16 = y_C^2 - 8y_C + 16\)

\(8y_C = -8y_C\)

\(16y_C = 0 \implies y_C = 0\)

Substitute \(y_C = 0\) back into Equation 1:

\(x_C^2 + (0 + 4)^2 = 64\)

\(x_C^2 + 16 = 64\)

\(x_C^2 = 48\)

\(x_C = \pm \sqrt{48} = \pm \sqrt{16 \times 3} = \pm 4\sqrt{3}\)

So, the coordinates of C can be \((4\sqrt{3}, 0)\) or \((-4\sqrt{3}, 0)\). For the resulting equation to match one of the options, let's consider \(C = (4\sqrt{3}, 0)\).

• A = (0, -4)
• B = (0, 4)
• C = \((4\sqrt{3}, 0)\)

Calculating the Slope of Side AC

The line we are interested in is parallel to side AC. First, let's find the slope of the line passing through A (0, -4) and C \((4\sqrt{3}, 0)\). The slope \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(m = \frac{y_2 - y_1}{x_2 - x_1}\).

Slope of AC \((m_{AC})\):

\(m_{AC} = \frac{0 - (-4)}{4\sqrt{3} - 0} = \frac{4}{4\sqrt{3}} = \frac{1}{\sqrt{3}}\)

Finding the Equation of the Parallel Line

We need the equation of a line that passes through the point (8, 0) and is parallel to AC. Parallel lines have the same slope.

So, the slope of the required line is \(m = \frac{1}{\sqrt{3}}\).

We use the point-slope form of the equation of a line: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (8, 0)\) and \(m = \frac{1}{\sqrt{3}}\).

\(y - 0 = \frac{1}{\sqrt{3}}(x - 8)\)

\(y = \frac{1}{\sqrt{3}}(x - 8)\)

Multiply both sides by \(\sqrt{3}\) to eliminate the denominator:

\(\sqrt{3}y = x - 8\)

Rearrange the terms to get the equation in the form \(Ax + By + C = 0\):

\(x - \sqrt{3}y - 8 = 0\)

Comparing with the Options

Let's compare our derived equation with the given options:

• Option 1: \(x - \sqrt{3}y - 8 = 0\)
• Option 2: \(x + \sqrt{3}y - 8 = 0\)
• Option 3: \(\sqrt{3}x + y - 8\sqrt{3} = 0\)
• Option 4: \(\sqrt{3}x - y - 8\sqrt{3} = 0\)

Our equation \(x - \sqrt{3}y - 8 = 0\) matches Option 1.

Note: If we had chosen \(C = (-4\sqrt{3}, 0)\), the slope of AC would have been \(m_{AC} = \frac{0 - (-4)}{-4\sqrt{3} - 0} = \frac{4}{-4\sqrt{3}} = -\frac{1}{\sqrt{3}}\). The equation of the line through (8, 0) with this slope would be \(y - 0 = -\frac{1}{\sqrt{3}}(x - 8) \implies \sqrt{3}y = -(x - 8) \implies \sqrt{3}y = -x + 8 \implies x + \sqrt{3}y - 8 = 0\), which is Option 2. The problem could potentially have two valid interpretations for C's position, but Option 1 is presented as the correct answer, suggesting the case where C has a positive x-coordinate.

Revision Table: Key Concepts for Triangle and Line Equations

Additional Information: Equilateral Triangle Properties and Line Equations

An equilateral triangle is a special type of triangle where all three sides are equal in length, and all three interior angles are equal to 60 degrees. This symmetry allows us to use various geometric properties to find coordinates or lengths.

When working with coordinate geometry, understanding how to represent geometric shapes using coordinates is fundamental. The equations of lines are crucial tools:

• Slope-Intercept Form: \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept.
• Point-Slope Form: \(y - y_1 = m(x - x_1)\), useful when you know the slope and one point on the line.
• Standard Form: \(Ax + By + C = 0\), a general form for linear equations.

Parallel lines never intersect and have the same slope. Perpendicular lines (if not vertical/horizontal) have slopes whose product is -1 (\(m_1 \times m_2 = -1\)).

In this problem, the position of the triangle on the coordinate plane was specified precisely, allowing us to calculate the exact coordinates of its vertices. Finding the slope of AC was the key step to determining the slope of the parallel line, which then led to the final equation.