Let `2a+2b+c=0` , then the equation of the straight line ax + by + c = 0 which is farthest the point (1,1) is

Options

A. y = x
B. y + x = 2
C. y + x = 4
D. y = x + 2

Answer

To solve the problem, we need to find the equation of the straight line \( ax + by + c = 0 \) that is farthest from the point \( (1, 1) \) given the condition \( 2a + 2b + c = 0 \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
We start with the equation of the line \( ax + by + c = 0 \). The condition given is \( 2a + 2b + c = 0 \). We can express \( c \) in terms of \( a \) and \( b \):
\[ c = -2a - 2b \]
Thus, the equation of the line can be rewritten as:
\[ ax + by - 2a - 2b = 0 \]
or
\[ ax + by = 2a + 2b \]

2. **Distance from a Point to a Line**:
The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
In our case, \( A = a \), \( B = b \), and \( C = -2a - 2b \). The point is \( (1, 1) \).

3. **Substituting the Point**:
We substitute \( (1, 1) \) into the distance formula:
\[ d = \frac{|a(1) + b(1) - 2a - 2b|}{\sqrt{a^2 + b^2}} = \frac{|a + b - 2a - 2b|}{\sqrt{a^2 + b^2}} = \frac{| -a - b |}{\sqrt{a^2 + b^2}} = \frac{a + b}{\sqrt{a^2 + b^2}} \quad (\text{since } a + b \geq 0) \]

4. **Maximizing the Distance**:
To maximize the distance \( d \), we need to maximize \( \frac{a + b}{\sqrt{a^2 + b^2}} \). This can be done by using the Cauchy-Schwarz inequality:
\[ (a + b)^2 \leq (1^2 + 1^2)(a^2 + b^2) \implies (a + b)^2 \leq 2(a^2 + b^2) \]
Thus,
\[ \frac{a + b}{\sqrt{a^2 + b^2}} \leq \sqrt{2} \]

5. **Finding the Line**:
The maximum distance occurs when \( a = b \). Let \( a = k \) and \( b = k \):
\[ 2k + c = 0 \implies c = -2k \]
Therefore, the line equation becomes:
\[ kx + ky - 2k = 0 \implies x + y - 2 = 0 \implies x + y = 2 \]

### Final Equation:
The equation of the line that is farthest from the point \( (1, 1) \) is:
\[ x + y = 2 \]