The three lines 4x +4y = 1 , 8 x - 3y = 2 , y = 0 are

Options

A. The sides of an isosceles triangle
B. Concurrent
C. Mutually perpendicular
D. The sides of an equilateral triangle

Answer

To determine the nature of the three lines given by the equations \(4x + 4y = 1\), \(8x - 3y = 2\), and \(y = 0\), we will find their intersection points and analyze their relationships.

### Step 1: Identify the equations of the lines
The equations of the lines are:
1. \(4x + 4y = 1\) (Equation 1)
2. \(8x - 3y = 2\) (Equation 2)
3. \(y = 0\) (Equation 3)

### Step 2: Solve for the intersection of Equation 1 and Equation 3
Substituting \(y = 0\) into Equation 1:
\[ 4x + 4(0) = 1 \implies 4x = 1 \implies x = \frac{1}{4} \]
Thus, the intersection point is \(\left(\frac{1}{4}, 0\right)\).

### Step 3: Solve for the intersection of Equation 2 and Equation 3
Substituting \(y = 0\) into Equation 2:
\[ 8x - 3(0) = 2 \implies 8x = 2 \implies x = \frac{2}{8} = \frac{1}{4} \]
Thus, the intersection point is also \(\left(\frac{1}{4}, 0\right)\).

### Step 4: Conclusion about concurrency
Since both pairs of lines intersect at the same point \(\left(\frac{1}{4}, 0\right)\), we conclude that the three lines are concurrent.

### Step 5: Check for perpendicularity
To check if the lines are mutually perpendicular, we need to find the slopes of the lines.

1. **For Equation 1**: Rearranging \(4x + 4y = 1\) gives \(y = -x + \frac{1}{4}\). The slope \(m_1 = -1\).
2. **For Equation 2**: Rearranging \(8x - 3y = 2\) gives \(y = \frac{8}{3}x - \frac{2}{3}\). The slope \(m_2 = \frac{8}{3}\).
3. **For Equation 3**: The line \(y = 0\) is horizontal, so its slope \(m_3 = 0\).

To check if two lines are perpendicular, the product of their slopes should equal \(-1\):
- For lines 1 and 2: \(m_1 \cdot m_2 = -1 \cdot \frac{8}{3} \neq -1\) (not perpendicular).
- For lines 1 and 3: \(m_1 \cdot m_3 = -1 \cdot 0 = 0\) (not perpendicular).
- For lines 2 and 3: \(m_2 \cdot m_3 = \frac{8}{3} \cdot 0 = 0\) (not perpendicular).

Thus, the lines are not mutually perpendicular.

### Step 6: Check for equilateral triangle
To check if the lines form an equilateral triangle, we would need to find the lengths of the segments formed by the intersection points and check if they are all equal. However, since we only have one intersection point and the lines do not form a closed triangle, they cannot form an equilateral triangle.

### Final Conclusion
The three lines are concurrent but not mutually perpendicular and do not form an equilateral triangle.

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