In a triangle ABC , if the equation of sides AB,BC and CA are `2x- y + 4 = 0 , x - 2y - 1 = 0 and x + 3y - 3 = 0 ` respectively ,Tangent of internal angle A
is equal to

Options

A. (a) `-7`
B. (b) `-3`
C. (c) `1/2`
D. (d) 7

Answer

To find the tangent of the internal angle A in triangle ABC given the equations of its sides, we will follow these steps:

### Step 1: Identify the equations of the sides
The equations of the sides of triangle ABC are:
1. AB: \(2x - y + 4 = 0\)
2. BC: \(x - 2y - 1 = 0\)
3. CA: \(x + 3y - 3 = 0\)

### Step 2: Rearrange the equations to find the slopes
To find the slopes of the lines, we will rearrange each equation into the slope-intercept form \(y = mx + c\), where \(m\) is the slope.

**For line AB:**
\[ 2x - y + 4 = 0 \implies y = 2x + 4 \]
Thus, the slope \(m_1\) of line AB is \(2\).

**For line AC:**
\[ x + 3y - 3 = 0 \implies 3y = -x + 3 \implies y = -\frac{1}{3}x + 1 \]
Thus, the slope \(m_2\) of line AC is \(-\frac{1}{3}\).

### Step 3: Use the formula for tangent of the angle between two lines
The formula for the tangent of the angle \(A\) formed between two lines with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan A = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Since we are looking for the internal angle, we will use the formula as is.

### Step 4: Substitute the slopes into the formula
Substituting \(m_1 = 2\) and \(m_2 = -\frac{1}{3}\):
\[ \tan A = \frac{2 - \left(-\frac{1}{3}\right)}{1 + 2 \left(-\frac{1}{3}\right)} \]
This simplifies to:
\[ \tan A = \frac{2 + \frac{1}{3}}{1 - \frac{2}{3}} = \frac{\frac{6}{3} + \frac{1}{3}}{\frac{1}{3}} = \frac{\frac{7}{3}}{\frac{1}{3}} \]

### Step 5: Simplify the expression
\[ \tan A = \frac{7}{3} \cdot 3 = 7 \]

### Step 6: Determine the sign of the tangent
Since we are looking for the tangent of the internal angle A, we take the negative value:
\[ \tan A = -7 \]

### Conclusion
Thus, the tangent of the internal angle A is:
\[ \text{Answer: } -7 \]