Let `A=(a,b)` and `B=(c,d)` where `c gt a gt 0` and `d gt b gt 0`. Then, point C on the X-axis such that `AC + BC` is the minimum, is:

Options

A. `(bc-ad)/(b-d)`
B. `(ac+bd)/(b+d)`
C. `(ac-bd)/(b-a)`
D. `(ad+bc)/(b+d)`

Answer

To find the point \( C \) on the X-axis such that \( AC + BC \) is minimized, we can follow these steps:

### Step 1: Define the points
Let \( A = (a, b) \) and \( B = (c, d) \), where \( c > a > 0 \) and \( d > b > 0 \). The point \( C \) will have coordinates \( (x, 0) \) since it lies on the X-axis.

### Step 2: Write the distance expressions
The distances \( AC \) and \( BC \) can be expressed using the distance formula:
\[ AC = \sqrt{(x - a)^2 + (0 - b)^2} = \sqrt{(x - a)^2 + b^2} \]
\[ BC = \sqrt{(x - c)^2 + (0 - d)^2} = \sqrt{(x - c)^2 + d^2} \]

### Step 3: Set up the total distance function
We want to minimize the total distance \( AC + BC \):
\[ D(x) = AC + BC = \sqrt{(x - a)^2 + b^2} + \sqrt{(x - c)^2 + d^2} \]

### Step 4: Use reflection to simplify the problem
To minimize \( D(x) \), we can reflect point \( B \) across the X-axis to get point \( B' = (c, -d) \). The new distance we want to minimize is \( AC + CB' \).

### Step 5: Find the coordinates of point \( C \)
The minimum distance \( AC + CB' \) occurs when point \( C \) lies on the straight line connecting points \( A \) and \( B' \). The equation of the line passing through points \( A \) and \( B' \) can be found using the two-point form of the line equation.

### Step 6: Calculate the intersection with the X-axis
To find the X-coordinate of point \( C \), we can use the section formula or the concept of similar triangles. The X-coordinate of point \( C \) can be calculated as:
\[ x = \frac{a \cdot d + b \cdot c}{b + d} \]

### Step 7: Final coordinates of point \( C \)
Thus, the coordinates of point \( C \) on the X-axis that minimizes \( AC + BC \) is:
\[ C = \left( \frac{a \cdot d + b \cdot c}{b + d}, 0 \right) \]