Let A(3, 2) and B(6, 5) be two points and a point C (x,y) is chosen on the line y=x such that AC+BC is minimum, then the co-ordinates of C are:

Options

A. (1,1)
B. (2,2)
C. (3,3)
D. (4,4)

Answer

To find the coordinates of point C (x, y) on the line y = x such that the sum of distances AC + BC is minimized, we can follow these steps:

### Step 1: Identify the points A and B
We have:
- Point A: A(3, 2)
- Point B: B(6, 5)

### Step 2: Reflect point A across the line y = x
The reflection of point A(3, 2) across the line y = x will be A'(2, 3). This is because the coordinates are swapped when reflecting over the line y = x.

### Step 3: Set up the distance formula
We want to minimize the distance AC + BC. However, instead of calculating this directly, we can minimize the distance A'B, where A' is the reflection of A.

### Step 4: Calculate the distance A'B
Using the distance formula:
\[ A'B = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting A'(2, 3) and B(6, 5):
\[ A'B = \sqrt{(6 - 2)^2 + (5 - 3)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]

### Step 5: Find the equation of line A'B
The slope of line A'B is given by:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{6 - 2} = \frac{2}{4} = \frac{1}{2} \]
Using point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting point A'(2, 3):
\[ y - 3 = \frac{1}{2}(x - 2) \]
Rearranging gives:
\[ y = \frac{1}{2}x + 2 \]

### Step 6: Find the intersection of line A'B with line y = x
To find point C, we set the equations equal:
\[ x = \frac{1}{2}x + 2 \]
Solving for x:
\[ x - \frac{1}{2}x = 2 \implies \frac{1}{2}x = 2 \implies x = 4 \]
Since C lies on the line y = x, we have:
\[ y = 4 \]
Thus, the coordinates of point C are (4, 4).

### Final Answer
The coordinates of point C are (4, 4).

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