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 \( (1, 1) \) under the condition \( 2a + 2b + c = 0 \).

### Step 1: Understand the given condition
We are given the condition:
\[ 2a + 2b + c = 0 \]
This can be rearranged to express \( c \) in terms of \( a \) and \( b \):
\[ c = -2a - 2b \]

### Step 2: Substitute \( c \) into the line equation
Substituting \( c \) into the line equation \( ax + by + c = 0 \):
\[ ax + by - 2a - 2b = 0 \]
This simplifies to:
\[ ax + by = 2a + 2b \]

### Step 3: Identify the point on the line
We need to find a line that passes through the point \( (2, 2) \) since substituting \( x = 2 \) and \( y = 2 \) into our line equation gives:
\[ a(2) + b(2) + c = 0 \implies 2a + 2b + c = 0 \]
This confirms that the point \( (2, 2) \) lies on the line.

### Step 4: Find the slope of the line
To maximize the distance from the point \( (1, 1) \) to the line, we need to find the slope of the line that is perpendicular to the line segment connecting \( (1, 1) \) and \( (2, 2) \). The slope of the line segment \( PQ \) (where \( P = (1, 1) \) and \( Q = (2, 2) \)) is:
\[ \text{slope of } PQ = \frac{2 - 1}{2 - 1} = 1 \]
The slope of the line we are looking for (let's call it \( m_L \)) must satisfy:
\[ m_{PQ} \cdot m_L = -1 \implies 1 \cdot m_L = -1 \implies m_L = -1 \]

### Step 5: Write the equation of the line
Now that we have the slope \( m_L = -1 \) and a point \( (2, 2) \), we can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (2, 2) \) and \( m = -1 \):
\[ y - 2 = -1(x - 2) \]
This simplifies to:
\[ y - 2 = -x + 2 \implies y + x = 4 \]

### Final Equation
Thus, the equation of the required line is:
\[ x + y = 4 \]