The equations of sides `AB`, `BC` and `AC` of `DeltaABC` are respectively `y=x`, `y=0` and `4x+3y=12`, then find :
`(i)` length of perpendicular from `B` to `AC`
`(ii) /_BAC`.

Answer

To solve the problem, we need to find the following for triangle \( \Delta ABC \) with given equations of sides:

1. Length of the perpendicular from point \( B \) to line \( AC \).
2. Angle \( \angle BAC \).

### Step 1: Identify the Points of Intersection

**Given Equations:**
- Line \( AB: y = x \)
- Line \( BC: y = 0 \) (the x-axis)
- Line \( AC: 4x + 3y = 12 \)

**Finding Points:**
1. **Point \( A \)**: Intersection of lines \( AB \) and \( BC \):
- Set \( y = x \) in \( y = 0 \):
\[ x = 0 \implies A(0, 0) \]

2. **Point \( B \)**: Intersection of lines \( AB \) and \( AC \):
- Substitute \( y = x \) into \( 4x + 3y = 12 \):
\[ 4x + 3x = 12 \implies 7x = 12 \implies x = \frac{12}{7}, \quad y = \frac{12}{7} \]
Thus, \( B\left(\frac{12}{7}, \frac{12}{7}\right) \).

3. **Point \( C \)**: Intersection of lines \( BC \) and \( AC \):
- Set \( y = 0 \) in \( 4x + 3y = 12 \):
\[ 4x = 12 \implies x = 3 \implies C(3, 0) \]

### Step 2: Length of Perpendicular from Point \( B \) to Line \( AC \)

**Equation of Line \( AC \)**:
- Rearranging \( 4x + 3y - 12 = 0 \) gives us \( a = 4, b = 3, c = -12 \).

**Using the Perpendicular Distance Formula**:
\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
Where \( (x_1, y_1) = B\left(\frac{12}{7}, \frac{12}{7}\right) \).

**Calculating the Distance**:
\[ d = \frac{|4 \cdot \frac{12}{7} + 3 \cdot \frac{12}{7} - 12|}{\sqrt{4^2 + 3^2}} = \frac{| \frac{48}{7} + \frac{36}{7} - 12 |}{\sqrt{16 + 9}} = \frac{| \frac{84}{7} - \frac{84}{7} |}{5} = \frac{0}{5} = 0 \]
This indicates that point \( B \) lies on line \( AC \), hence the perpendicular distance is 0.

### Step 3: Finding Angle \( \angle BAC \)

**Finding Slopes**:
1. **Slope of Line \( AC \)**:
- From \( 4x + 3y - 12 = 0 \), slope \( m_1 = -\frac{4}{3} \).

2. **Slope of Line \( AB \)**:
- From \( y = x \), slope \( m_2 = 1 \).

**Using the Formula for Angle Between Two Lines**:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting \( m_1 = -\frac{4}{3} \) and \( m_2 = 1 \):
\[ \tan(\theta) = \left| \frac{-\frac{4}{3} - 1}{1 + (-\frac{4}{3})(1)} \right| = \left| \frac{-\frac{4}{3} - \frac{3}{3}}{1 - \frac{4}{3}} \right| = \left| \frac{-\frac{7}{3}}{-\frac{1}{3}} \right| = 7 \]
Thus, \( \theta = \tan^{-1}(7) \).

### Final Answers:
1. Length of perpendicular from \( B \) to \( AC \): **0**
2. Angle \( \angle BAC \): **\( \tan^{-1}(7) \)**