A is the point (-1, 0) and B is the point (1, 1). Find a point on the line `4x+5y=4`, which is equidistant from A and B.

Answer

To find a point on the line \(4x + 5y = 4\) that is equidistant from points \(A(-1, 0)\) and \(B(1, 1)\), we can follow these steps:

### Step 1: Understand the Problem
We need to find a point \(C(x, y)\) on the line \(4x + 5y = 4\) such that the distance from \(C\) to \(A\) is equal to the distance from \(C\) to \(B\).

### Step 2: Write the Distance Formula
The distance \(AC\) from point \(A(-1, 0)\) to point \(C(x, y)\) is given by:
\[ AC = \sqrt{(x + 1)^2 + (y - 0)^2} \]
The distance \(BC\) from point \(B(1, 1)\) to point \(C(x, y)\) is given by:
\[ BC = \sqrt{(x - 1)^2 + (y - 1)^2} \]

### Step 3: Set the Distances Equal
Since \(C\) is equidistant from \(A\) and \(B\), we can set the distances equal:
\[ \sqrt{(x + 1)^2 + y^2} = \sqrt{(x - 1)^2 + (y - 1)^2} \]

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

### Step 5: Expand Both Sides
Expanding both sides gives:
\[ x^2 + 2x + 1 + y^2 = x^2 - 2x + 1 + y^2 - 2y + 1 \]
This simplifies to:
\[ x^2 + 2x + 1 + y^2 = x^2 - 2x + 2 + y^2 - 2y \]

### Step 6: Cancel Common Terms
Cancel \(x^2\) and \(y^2\) from both sides:
\[ 2x + 1 = -2x + 2 - 2y \]

### Step 7: Rearrange the Equation
Rearranging gives:
\[ 4x + 2y = 1 \]

### Step 8: Solve for y
From \(4x + 2y = 1\), we can express \(y\) in terms of \(x\):
\[ 2y = 1 - 4x \implies y = \frac{1 - 4x}{2} \]

### Step 9: Substitute into the Line Equation
Now we substitute \(y\) into the line equation \(4x + 5y = 4\):
\[ 4x + 5\left(\frac{1 - 4x}{2}\right) = 4 \]

### Step 10: Simplify the Equation
Multiplying through by 2 to eliminate the fraction:
\[ 8x + 5(1 - 4x) = 8 \]
Expanding gives:
\[ 8x + 5 - 20x = 8 \]
Combining like terms results in:
\[ -12x + 5 = 8 \]

### Step 11: Solve for x
Rearranging gives:
\[ -12x = 3 \implies x = -\frac{1}{4} \]

### Step 12: Find y
Substituting \(x = -\frac{1}{4}\) back into \(y = \frac{1 - 4x}{2}\):
\[ y = \frac{1 - 4\left(-\frac{1}{4}\right)}{2} = \frac{1 + 1}{2} = 1 \]

### Step 13: Conclusion
Thus, the point \(C\) is:
\[ C\left(-\frac{1}{4}, 1\right) \]

### Final Answer
The point on the line \(4x + 5y = 4\) that is equidistant from points \(A\) and \(B\) is \(\left(-\frac{1}{4}, 1\right)\).