To find the value of 'a' given the points A (-3, -2), B (-6, a), C (-3, -4), and D (0, -1) such that AB = CD and 'a' is negative, we will use the distance formula.
### Step-by-Step Solution:
1. **Write the Distance Formula**:
The distance between two points (x1, y1) and (x2, y2) is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
2. **Calculate AB**:
For points A (-3, -2) and B (-6, a):
- \(x_1 = -3\), \(y_1 = -2\)
- \(x_2 = -6\), \(y_2 = a\)
\[
AB = \sqrt{((-6) - (-3))^2 + (a - (-2))^2}
\]
\[
AB = \sqrt{(-6 + 3)^2 + (a + 2)^2}
\]
\[
AB = \sqrt{(-3)^2 + (a + 2)^2}
\]
\[
AB = \sqrt{9 + (a + 2)^2}
\]
3. **Calculate CD**:
For points C (-3, -4) and D (0, -1):
- \(x_1 = -3\), \(y_1 = -4\)
- \(x_2 = 0\), \(y_2 = -1\)
\[
CD = \sqrt{(0 - (-3))^2 + (-1 - (-4))^2}
\]
\[
CD = \sqrt{(0 + 3)^2 + (-1 + 4)^2}
\]
\[
CD = \sqrt{3^2 + 3^2}
\]
\[
CD = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
\]
4. **Set AB equal to CD**:
Since we know \(AB = CD\):
\[
\sqrt{9 + (a + 2)^2} = 3\sqrt{2}
\]
5. **Square both sides**:
To eliminate the square root:
\[
9 + (a + 2)^2 = (3\sqrt{2})^2
\]
\[
9 + (a + 2)^2 = 18
\]
6. **Solve for (a + 2)^2**:
\[
(a + 2)^2 = 18 - 9
\]
\[
(a + 2)^2 = 9
\]
7. **Take the square root**:
\[
a + 2 = \pm 3
\]
This gives us two equations:
- \(a + 2 = 3\)
- \(a + 2 = -3\)
8. **Solve for 'a'**:
- From \(a + 2 = 3\):
\[
a = 3 - 2 = 1
\]
- From \(a + 2 = -3\):
\[
a = -3 - 2 = -5
\]
9. **Select the negative solution**:
Since it is given that 'a' is negative, we choose:
\[
a = -5
\]
### Final Answer:
Thus, the value of 'a' is \(-5\).
