Points on the line `x+y=4` that lie at a unit distance from the line `4x+3y-10=0` are

Options

A. (3,1) and (-7,11)
B. (-3,7) and (2,2)
C. (-3,7) and (-7,11)
D. None of these

Answer

To find the points on the line \(x + y = 4\) that lie at a unit distance from the line \(4x + 3y - 10 = 0\), we will follow these steps:

### Step 1: Identify the equations of the lines
The first line is given by:
\[ x + y = 4 \]
The second line is given by:
\[ 4x + 3y - 10 = 0 \]

### Step 2: Find the slope and intercept of the second line
Rearranging the second line into slope-intercept form \(y = mx + b\):
\[ 3y = -4x + 10 \implies y = -\frac{4}{3}x + \frac{10}{3} \]
The slope of the line \(4x + 3y - 10 = 0\) is \(-\frac{4}{3}\).

### Step 3: Calculate the distance from a point to the line
The formula for the distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For our line \(4x + 3y - 10 = 0\), we have \(A = 4\), \(B = 3\), and \(C = -10\).

### Step 4: Set up the distance equation
We need the distance to be equal to 1 unit:
\[ \frac{|4x_1 + 3y_1 - 10|}{\sqrt{4^2 + 3^2}} = 1 \]
Calculating the denominator:
\[ \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
Thus, we have:
\[ |4x_1 + 3y_1 - 10| = 5 \]

### Step 5: Solve the absolute value equation
This gives us two equations:
1. \(4x_1 + 3y_1 - 10 = 5\)
2. \(4x_1 + 3y_1 - 10 = -5\)

#### For the first equation:
\[ 4x_1 + 3y_1 = 15 \]

#### For the second equation:
\[ 4x_1 + 3y_1 = 5 \]

### Step 6: Substitute \(y_1\) from the first line equation
From the line \(x + y = 4\), we can express \(y_1\) as:
\[ y_1 = 4 - x_1 \]

### Step 7: Substitute into the equations
Substituting \(y_1\) into the first equation:
\[ 4x_1 + 3(4 - x_1) = 15 \]
\[ 4x_1 + 12 - 3x_1 = 15 \implies x_1 + 12 = 15 \implies x_1 = 3 \]
Then substituting back to find \(y_1\):
\[ y_1 = 4 - 3 = 1 \]
So one point is \((3, 1)\).

Now substituting into the second equation:
\[ 4x_1 + 3(4 - x_1) = 5 \]
\[ 4x_1 + 12 - 3x_1 = 5 \implies x_1 + 12 = 5 \implies x_1 = -7 \]
Then substituting back to find \(y_1\):
\[ y_1 = 4 - (-7) = 11 \]
So the second point is \((-7, 11)\).

### Step 8: Conclusion
The points on the line \(x + y = 4\) that lie at a unit distance from the line \(4x + 3y - 10 = 0\) are:
\[ (3, 1) \text{ and } (-7, 11) \]