The two points on the line `x+y=4` that lies at a unit perpendicular distance from the line `4x+3y=10` are `(a_(1), b_(1))` and `(a_(2), b_(2))` then `a_(1)+b_(1)+a_(2)+b_(2)` is equal to (a) 5 (b) 6 (c) 7 (d) 8

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the problem, we need to find two points on the line \(x + y = 4\) that are at a unit perpendicular distance from the line \(4x + 3y = 10\). We will denote these points as \((a_1, b_1)\) and \((a_2, b_2)\) and then calculate \(a_1 + b_1 + a_2 + b_2\).

### Step 1: Express the points on the line \(x + y = 4\)

Any point on the line \(x + y = 4\) can be expressed in terms of \(x\):
\[ y = 4 - x \]
Thus, a point on the line can be represented as \((x_1, 4 - x_1)\).

### Step 2: Find the distance from the point to the line

The distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is given by the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(4x + 3y = 10\), we can rewrite it in the form \(Ax + By + C = 0\):
\[ 4x + 3y - 10 = 0 \quad \Rightarrow \quad A = 4, B = 3, C = -10 \]

### Step 3: Set the distance equal to 1

Substituting \(x_1\) and \(y_1\) into the distance formula, we have:
\[ d = \frac{|4x_1 + 3(4 - x_1) - 10|}{\sqrt{4^2 + 3^2}} = 1 \]
Calculating the denominator:
\[ \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]
Thus, the equation becomes:
\[ \frac{|4x_1 + 12 - 3x_1 - 10|}{5} = 1 \]
Simplifying the numerator:
\[ \frac{|x_1 + 2|}{5} = 1 \]
Multiplying both sides by 5 gives:
\[ |x_1 + 2| = 5 \]

### Step 4: Solve for \(x_1\)

This absolute value equation gives us two cases:
1. \(x_1 + 2 = 5\)
2. \(x_1 + 2 = -5\)

From the first case:
\[ x_1 = 5 - 2 = 3 \]
From the second case:
\[ x_1 = -5 - 2 = -7 \]

### Step 5: Find corresponding \(y_1\) values

Now we find \(y_1\) for both \(x_1\) values:
- For \(x_1 = 3\):
\[ y_1 = 4 - 3 = 1 \quad \Rightarrow \quad (a_1, b_1) = (3, 1) \]
- For \(x_1 = -7\):
\[ y_1 = 4 - (-7) = 4 + 7 = 11 \quad \Rightarrow \quad (a_2, b_2) = (-7, 11) \]

### Step 6: Calculate \(a_1 + b_1 + a_2 + b_2\)

Now we can calculate:
\[ a_1 + b_1 + a_2 + b_2 = 3 + 1 + (-7) + 11 \]
Calculating step by step:
\[ 3 + 1 = 4 \]
\[ 4 - 7 = -3 \]
\[ -3 + 11 = 8 \]

### Final Result

Thus, the value of \(a_1 + b_1 + a_2 + b_2\) is:
\[ \boxed{8} \]