The points A(-1,2),B(x,y)and C = (4,5) are such that BA = BC . Find a linear relations between x and y .

Answer

To solve the problem where the points A(-1,2), B(x,y), and C(4,5) are such that the distance from B to A is equal to the distance from B to C (BA = BC), we will follow these steps:

### Step 1: Use the Distance Formula
The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

### Step 2: Calculate Distance BA
For points A(-1, 2) and B(x, y), the distance BA is:
\[ BA = \sqrt{(x - (-1))^2 + (y - 2)^2} = \sqrt{(x + 1)^2 + (y - 2)^2} \]

### Step 3: Calculate Distance BC
For points B(x, y) and C(4, 5), the distance BC is:
\[ BC = \sqrt{(4 - x)^2 + (5 - y)^2} \]

### Step 4: Set Distances Equal
Since BA = BC, we set the two distances equal:
\[ \sqrt{(x + 1)^2 + (y - 2)^2} = \sqrt{(4 - x)^2 + (5 - y)^2} \]

### Step 5: Square Both Sides
To eliminate the square roots, we square both sides:
\[ (x + 1)^2 + (y - 2)^2 = (4 - x)^2 + (5 - y)^2 \]

### Step 6: Expand Both Sides
Now we will expand both sides:
- Left side:
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
So,
\[ x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 + 2x - 4y + 5 \]

- Right side:
\[ (4 - x)^2 = 16 - 8x + x^2 \]
\[ (5 - y)^2 = 25 - 10y + y^2 \]
So,
\[ 16 - 8x + x^2 + 25 - 10y + y^2 = x^2 + y^2 - 8x - 10y + 41 \]

### Step 7: Set the Expanded Equations Equal
Now we have:
\[ x^2 + 2x - 4y + 5 = x^2 + y^2 - 8x - 10y + 41 \]

### Step 8: Cancel \(x^2\) and \(y^2\) and Rearrange
Cancel \(x^2\) and \(y^2\) from both sides:
\[ 2x - 4y + 5 = -8x - 10y + 41 \]
Rearranging gives:
\[ 2x + 8x + 10y - 4y = 41 - 5 \]
\[ 10x + 6y = 36 \]

### Step 9: Simplify the Equation
Dividing the entire equation by 2 gives:
\[ 5x + 3y = 18 \]

### Final Result
The linear relation between \(x\) and \(y\) is:
\[ 5x + 3y = 18 \]