Equation of the base of an equilateral triangle is `3x + 4y = 9 ` and its vertex is at point (1,2) .Find the equations of the other sides and the length of each side of the triangle .

Answer

To solve the problem, we need to find the equations of the other sides of the equilateral triangle and the length of each side. Here’s a step-by-step solution:

### Step 1: Identify the given information
We have the equation of the base of the equilateral triangle:
\[ 3x + 4y = 9 \]
And the vertex of the triangle is at the point:
\[ A(1, 2) \]

### Step 2: Find the slope of the base
To find the slope of the line given by the equation \(3x + 4y = 9\), we can rewrite it in slope-intercept form \(y = mx + b\):
\[ 4y = -3x + 9 \implies y = -\frac{3}{4}x + \frac{9}{4} \]
Thus, the slope \(m_1\) of the base \(BC\) is:
\[ m_1 = -\frac{3}{4} \]

### Step 3: Determine the angles for the other sides
Since the triangle is equilateral, the angles between the sides \(AB\) and \(AC\) with the base \(BC\) are \(60^\circ\). The slopes of lines that make a \(60^\circ\) angle with a given slope can be found using the tangent of the angle.

### Step 4: Calculate the slopes of lines \(AB\) and \(AC\)
Using the formula for the tangent of the angle between two lines:
\[ \tan \theta = \frac{m_2 - m_1}{1 + m_1 m_2} \]
For \(60^\circ\), \(\tan 60^\circ = \sqrt{3}\). Setting \(m_2\) as the slope of \(AB\) or \(AC\):
\[ \sqrt{3} = \frac{m + \frac{3}{4}}{1 - \frac{3}{4}m} \]
Solving for \(m\) gives us two possible slopes for \(AB\) and \(AC\).

1. **Positive case**:
\[ \sqrt{3} = \frac{m + \frac{3}{4}}{1 - \frac{3}{4}m} \]
Cross-multiplying and simplifying will yield:
\[ m = \frac{4\sqrt{3} - 3}{4 + 3\sqrt{3}} \]

2. **Negative case**:
\[ -\sqrt{3} = \frac{m + \frac{3}{4}}{1 - \frac{3}{4}m} \]
Similarly, solving this will yield:
\[ m = \frac{4\sqrt{3} + 3}{3\sqrt{3} - 4} \]

### Step 5: Find the equations of lines \(AB\) and \(AC\)
Using the point-slope form of the line equation \(y - y_1 = m(x - x_1)\):

1. For slope \(m_1\):
\[ y - 2 = \left(\frac{4\sqrt{3} - 3}{4 + 3\sqrt{3}}\right)(x - 1) \]

2. For slope \(m_2\):
\[ y - 2 = \left(\frac{4\sqrt{3} + 3}{3\sqrt{3} - 4}\right)(x - 1) \]

### Step 6: Calculate the length of each side
To find the length of each side, we need the perpendicular distance from point \(A(1,2)\) to the line \(BC\):
Using the formula for the distance from a point to a line \(Ax + By + C = 0\):
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(3x + 4y - 9 = 0\):
\[ d = \frac{|3(1) + 4(2) - 9|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 8 - 9|}{\sqrt{9 + 16}} = \frac{2}{5} \]

Using the sine of \(60^\circ\):
\[ \sin 60^\circ = \frac{\text{perpendicular}}{\text{side length}} \implies \frac{\sqrt{3}}{2} = \frac{\frac{2}{5}}{A} \]
Solving for \(A\):
\[ A = \frac{2}{5} \cdot \frac{2}{\sqrt{3}} = \frac{4\sqrt{3}}{15} \]

### Final Answer
The equations of the other sides \(AB\) and \(AC\) can be derived from the slopes calculated, and the length of each side of the triangle is:
\[ \text{Length} = \frac{4\sqrt{3}}{15} \]