`x` coordinates of two points B and C are the roots of equation `x^2 +4x+3=0` and their `y` coordinates are the roots of equation `x^2 -x-6=0`. If `x` coordinate of B is less than the `x` coordinate of C and `y` coordinate of B is greater than the `y` coordinate of C and coordinates of a third point A be `(3, -5)`, find the length of the bisector of the interior angle at A.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Find the x-coordinates of points B and C

We need to solve the quadratic equation \(x^2 + 4x + 3 = 0\).

Using the factorization method:
\[ x^2 + 4x + 3 = (x + 1)(x + 3) = 0 \]
Thus, the roots are:
\[ x = -1 \quad \text{and} \quad x = -3 \]
Since the x-coordinate of B is less than that of C, we assign:
\[ B(-3, y_B) \quad \text{and} \quad C(-1, y_C) \]

### Step 2: Find the y-coordinates of points B and C

Now, we solve the quadratic equation \(x^2 - x - 6 = 0\).

Using the factorization method:
\[ x^2 - x - 6 = (x - 3)(x + 2) = 0 \]
Thus, the roots are:
\[ y = 3 \quad \text{and} \quad y = -2 \]
Since the y-coordinate of B is greater than that of C, we assign:
\[ B(-3, 3) \quad \text{and} \quad C(-1, -2) \]

### Step 3: Find the length of AB

Given point A is \(A(3, -5)\), we will use the distance formula to find the length of AB:
\[ AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \]
Substituting the coordinates:
\[ AB = \sqrt{(-3 - 3)^2 + (3 - (-5))^2} \]
\[ = \sqrt{(-6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]

### Step 4: Find the length of AC

Now we find the length of AC:
\[ AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} \]
Substituting the coordinates:
\[ AC = \sqrt{(-1 - 3)^2 + (-2 - (-5))^2} \]
\[ = \sqrt{(-4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 5: Use the Angle Bisector Theorem

According to the Angle Bisector Theorem:
\[ \frac{BD}{DC} = \frac{AB}{AC} = \frac{10}{5} = 2 \]
This means that \(BD:DC = 2:1\).

### Step 6: Find the coordinates of point D using the section formula

Using the section formula, where D divides BC in the ratio \(2:1\):
\[ D\left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right) \]
Here, \(m_1 = 2\), \(m_2 = 1\), \(B(-3, 3)\), and \(C(-1, -2)\):
\[ D\left(\frac{2 \cdot (-1) + 1 \cdot (-3)}{2 + 1}, \frac{2 \cdot (-2) + 1 \cdot 3}{2 + 1}\right) \]
Calculating the coordinates:
\[ D\left(\frac{-2 - 3}{3}, \frac{-4 + 3}{3}\right) = D\left(\frac{-5}{3}, \frac{-1}{3}\right) \]

### Step 7: Find the length of AD

Now we find the length of AD:
\[ AD = \sqrt{(x_D - x_A)^2 + (y_D - y_A)^2} \]
Substituting the coordinates:
\[ AD = \sqrt{\left(3 - \frac{-5}{3}\right)^2 + \left(-5 - \frac{-1}{3}\right)^2} \]
Calculating:
\[ = \sqrt{\left(3 + \frac{5}{3}\right)^2 + \left(-5 + \frac{1}{3}\right)^2} \]
\[ = \sqrt{\left(\frac{9}{3} + \frac{5}{3}\right)^2 + \left(-\frac{15}{3} + \frac{1}{3}\right)^2} \]
\[ = \sqrt{\left(\frac{14}{3}\right)^2 + \left(-\frac{14}{3}\right)^2} \]
\[ = \sqrt{\frac{196}{9} + \frac{196}{9}} = \sqrt{\frac{392}{9}} = \frac{14\sqrt{2}}{3} \]

### Final Answer
The length of the bisector of the interior angle at A is \(\frac{14\sqrt{2}}{3}\).

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