Understanding Concurrency of Lines

The question asks about the geometric relationship between three given lines: \(4x + 4y = 1\), \(8x - 3y = 2\), and \(y = 0\). We need to determine if they form a specific type of triangle, are concurrent, or are mutually perpendicular.

What does it mean for lines to be concurrent?

Three or more lines are said to be concurrent if they all intersect at a single point. If they are not concurrent, they might intersect at multiple points (forming a triangle) or some might be parallel.

Checking for Concurrency

A common method to check if three lines are concurrent is to find the intersection point of any two of the lines and then verify if this point lies on the third line.

Let's label the given lines:

• Line 1: \(4x + 4y = 1\)
• Line 2: \(8x - 3y = 2\)
• Line 3: \(y = 0\)

The equation of Line 3, \(y = 0\), is very simple, representing the x-axis. It is strategic to use this line when finding intersection points.

Finding the Intersection of Line 1 and Line 3

Substitute \(y = 0\) into the equation for Line 1:

\(4x + 4(0) = 1\)

\(4x + 0 = 1\)

\(4x = 1\)

\(x = \frac{1}{4}\)

So, the intersection point of Line 1 and Line 3 is \((\frac{1}{4}, 0)\).

Finding the Intersection of Line 2 and Line 3

Substitute \(y = 0\) into the equation for Line 2:

\(8x - 3(0) = 2\)

\(8x - 0 = 2\)

\(8x = 2\)

\(x = \frac{2}{8}\)

\(x = \frac{1}{4}\)

So, the intersection point of Line 2 and Line 3 is also \((\frac{1}{4}, 0)\).

Conclusion on Concurrency

We found that the intersection point of Line 1 and Line 3 is \((\frac{1}{4}, 0)\), and the intersection point of Line 2 and Line 3 is also \((\frac{1}{4}, 0)\). This means that both Line 1 and Line 2 pass through the point \((\frac{1}{4}, 0)\) which is on Line 3 (\(y = 0\)).

Since all three lines intersect at the single point \((\frac{1}{4}, 0)\), the lines are concurrent.

Examining Other Options

Are the lines mutually perpendicular?

For lines to be mutually perpendicular, each pair of lines must be perpendicular. The slopes of the lines are:

• Line 1 (\(4x + 4y = 1\)): \(4y = -4x + 1 \implies y = -x + \frac{1}{4}\). Slope \(m_1 = -1\).
• Line 2 (\(8x - 3y = 2\)): \(-3y = -8x + 2 \implies y = \frac{8}{3}x - \frac{2}{3}\). Slope \(m_2 = \frac{8}{3}\).
• Line 3 (\(y = 0\)): This is a horizontal line. Slope \(m_3 = 0\).

Checking perpendicularity (product of slopes is -1):

• \(m_1 \times m_2 = -1 \times \frac{8}{3} = -\frac{8}{3} \neq -1\) (Line 1 and Line 2 are not perpendicular)
• \(m_1 \times m_3 = -1 \times 0 = 0 \neq -1\) (Line 1 and Line 3 are not perpendicular)
• \(m_2 \times m_3 = \frac{8}{3} \times 0 = 0 \neq -1\) (Line 2 and Line 3 are not perpendicular, Line 3 is perpendicular only to vertical lines)

Since none of the pairs are perpendicular, the lines are not mutually perpendicular.

Are the lines the sides of an isosceles or equilateral triangle?

If the lines were the sides of a triangle, they would intersect at three distinct points. Since we found they intersect at a single point, they do not form a triangle. Therefore, they cannot be the sides of an isosceles or equilateral triangle.

Summary of Findings

By finding the intersection points, we confirmed that all three lines \(4x + 4y = 1\), \(8x - 3y = 2\), and \(y = 0\) intersect at the same point \((\frac{1}{4}, 0)\).

This confirms that the lines are concurrent.

Revision Table: Geometric Properties of Lines

Additional Information: The Determinant Method for Concurrency

For three lines given in the form \(a_ix + b_iy + c_i = 0\), they are concurrent if the determinant of the matrix formed by their coefficients is zero. The lines are \(4x + 4y - 1 = 0\), \(8x - 3y - 2 = 0\), and \(0x + 1y + 0 = 0\).

The determinant is:

\[ \begin{vmatrix} 4 & 4 & -1 \\ 8 & -3 & -2 \\ 0 & 1 & 0 \end{vmatrix} \]

Expanding along the third row:

\(0 \times (\text{cofactor of 0}) - 1 \times \begin{vmatrix} 4 & -1 \\ 8 & -2 \end{vmatrix} + 0 \times (\text{cofactor of 0})\)

\(= -1 \times ((4)(-2) - (-1)(8))\)

\(= -1 \times (-8 - (-8))\)

\(= -1 \times (-8 + 8)\)

\(= -1 \times 0\)

\(= 0\)

Since the determinant is 0, the lines are concurrent. This method provides an alternative way to confirm concurrency without finding the intersection point explicitly, especially useful for more complex equations.