If the line segment joining the points A(a,b) and B(c,d) subtends an angle `theta` at the origin, then `costheta` is equal to

Options

A. `(ab+cd)/(sqrt((a^(2)+b^(2))(c^(2)+d^(2))))`
B. `(ac+bd)/(sqrt((a^(2)+b^(2))(c^(2)+d^(2))))`
C. `(ac-bd)/(sqrt((a^(2)+b^(2))(c^(2)+d^(2))))`
D. `(ad+bc)/(sqrt((a^(2)+c^(2))(b^(2)+d^(2))))`

Answer

To solve the problem, we need to find the value of \(\cos \theta\) where \(\theta\) is the angle subtended at the origin by the line segment joining the points \(A(a, b)\) and \(B(c, d)\).

### Step-by-step Solution:

1. **Identify Points and Distances:**
- Let \(O\) be the origin \((0, 0)\), \(A(a, b)\), and \(B(c, d)\).
- We need to find the lengths of the segments \(OA\), \(OB\), and \(AB\).

2. **Calculate Distances:**
- The distance \(OA\) from the origin to point \(A\) is given by:
\[ OA = \sqrt{a^2 + b^2} \]
- The distance \(OB\) from the origin to point \(B\) is given by:
\[ OB = \sqrt{c^2 + d^2} \]
- The distance \(AB\) between points \(A\) and \(B\) is given by:
\[ AB = \sqrt{(c - a)^2 + (d - b)^2} \]

3. **Use the Cosine Rule:**
- According to the cosine rule in triangle \(OAB\):
\[ \cos \theta = \frac{OA^2 + OB^2 - AB^2}{2 \cdot OA \cdot OB} \]

4. **Substitute the Values:**
- Substitute the distances into the cosine formula:
\[ \cos \theta = \frac{(OA^2) + (OB^2) - (AB^2)}{2 \cdot OA \cdot OB} \]
- This becomes:
\[ \cos \theta = \frac{(a^2 + b^2) + (c^2 + d^2) - ((c - a)^2 + (d - b)^2)}{2 \cdot \sqrt{a^2 + b^2} \cdot \sqrt{c^2 + d^2}} \]

5. **Expand and Simplify:**
- Expand \((c - a)^2 + (d - b)^2\):
\[ (c - a)^2 + (d - b)^2 = (c^2 - 2ac + a^2) + (d^2 - 2bd + b^2) = c^2 + d^2 + a^2 + b^2 - 2ac - 2bd \]
- Substitute back into the equation:
\[ \cos \theta = \frac{(a^2 + b^2) + (c^2 + d^2) - (c^2 + d^2 + a^2 + b^2 - 2ac - 2bd)}{2 \cdot \sqrt{a^2 + b^2} \cdot \sqrt{c^2 + d^2}} \]
- Simplifying gives:
\[ \cos \theta = \frac{2ac + 2bd}{2 \cdot \sqrt{a^2 + b^2} \cdot \sqrt{c^2 + d^2}} \]

6. **Final Result:**
- Thus, we arrive at:
\[ \cos \theta = \frac{ac + bd}{\sqrt{a^2 + b^2} \cdot \sqrt{c^2 + d^2}} \]