A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2,0), (0,2) and (1,1) on the line is zero. Find the coordinate of the point P.

Explanation

To solve the problem, we need to find the coordinates of the fixed point P through which a variable line passes, given that the algebraic sum of the perpendiculars from the points (2,0), (0,2), and (1,1) on the line is zero.

### Step-by-Step Solution:

1. **Define the Fixed Point P**:
Let the coordinates of the fixed point P be (a, b).

2. **Equation of the Variable Line**:
The equation of the line passing through point P can be expressed in point-slope form:
\[ y - b = m(x - a) \]
Rearranging gives:
\[ mx - y + (b - ma) = 0 \]
This can be rewritten as:
\[ mx - y + (b - ma) = 0 \]

3. **Calculate the Perpendicular Distances**:
The perpendicular distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \(mx - y + (b - ma) = 0\):
- \(A = m\), \(B = -1\), \(C = b - ma\)

Now, we calculate the distances from the three points:

- **From (2, 0)**:
\[ d_1 = \frac{|m(2) - 0 + (b - ma)|}{\sqrt{m^2 + 1}} = \frac{|2m + b - ma|}{\sqrt{m^2 + 1}} \]

- **From (0, 2)**:
\[ d_2 = \frac{|m(0) - 2 + (b - ma)|}{\sqrt{m^2 + 1}} = \frac{|b - ma - 2|}{\sqrt{m^2 + 1}} \]

- **From (1, 1)**:
\[ d_3 = \frac{|m(1) - 1 + (b - ma)|}{\sqrt{m^2 + 1}} = \frac{|m - 1 + b - ma|}{\sqrt{m^2 + 1}} \]

4. **Set Up the Equation**:
According to the problem, the algebraic sum of these distances is zero:
\[ d_1 + d_2 + d_3 = 0 \]
This implies:
\[ \frac{|2m + b - ma| + |b - ma - 2| + |m - 1 + b - ma|}{\sqrt{m^2 + 1}} = 0 \]
Since the denominator is always positive, the numerator must be zero:
\[ |2m + b - ma| + |b - ma - 2| + |m - 1 + b - ma| = 0 \]

5. **Analyze the Absolute Values**:
Since the sum of absolute values is zero, each individual absolute value must also be zero:
- \(2m + b - ma = 0\)
- \(b - ma - 2 = 0\)
- \(m - 1 + b - ma = 0\)

6. **Solve the System of Equations**:
From \(b - ma - 2 = 0\), we get:
\[ b = ma + 2 \]
Substitute \(b\) into the other two equations:
- From \(2m + (ma + 2) - ma = 0\):
\[ 2m + 2 = 0 \implies m = -1 \]
- Substitute \(m = -1\) into \(b = ma + 2\):
\[ b = -a + 2 \]

7. **Find the Coordinates of P**:
Substitute \(m = -1\) into \(m - 1 + b - ma = 0\):
\[ -1 - 1 + b + a = 0 \implies b + a = 2 \]
Now we have two equations:
- \(b = -a + 2\)
- \(b + a = 2\)

Substituting \(b\) from the first equation into the second:
\[ -a + 2 + a = 2 \implies 2 = 2 \]
This is always true, indicating that \(a\) and \(b\) can take on values that satisfy both equations.

If we let \(a = 1\), then:
\[ b = -1 + 2 = 1 \]
Thus, the coordinates of point P are:
\[ P(1, 1) \]

### Final Answer:
The coordinates of the fixed point P are \((1, 1)\).