If the point A (-2, 1 ), B (a, b) and C (4, -1) are collinear and a-b = 1, find the values of a and b.

Answer

To find the values of \( a \) and \( b \) such that the points \( A(-2, 1) \), \( B(a, b) \), and \( C(4, -1) \) are collinear and \( a - b = 1 \), we will follow these steps:

### Step 1: Find the equation of the line through points A and C.
We can use the two-point form of the equation of a line, which is given by:

\[ \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \]

Here, \( A(-2, 1) \) is \( (x_1, y_1) \) and \( C(4, -1) \) is \( (x_2, y_2) \).

Substituting the coordinates:

\[ \frac{y - 1}{-1 - 1} = \frac{x + 2}{4 + 2} \]

This simplifies to:

\[ \frac{y - 1}{-2} = \frac{x + 2}{6} \]

### Step 2: Cross-multiply to eliminate the fractions.
Cross-multiplying gives:

\[ 6(y - 1) = -2(x + 2) \]

Expanding both sides:

\[ 6y - 6 = -2x - 4 \]

### Step 3: Rearranging the equation.
Rearranging the equation gives:

\[ 2x + 6y - 2 = 0 \]

Dividing the entire equation by 2 simplifies it to:

\[ x + 3y - 1 = 0 \]

### Step 4: Substitute point B into the line equation.
Since point \( B(a, b) \) lies on this line, we substitute \( a \) and \( b \) into the equation:

\[ a + 3b - 1 = 0 \]

This can be rearranged to form our first equation:

\[ a + 3b = 1 \quad \text{(Equation 1)} \]

### Step 5: Use the given condition \( a - b = 1 \).
We also have the condition given in the problem:

\[ a - b = 1 \quad \text{(Equation 2)} \]

### Step 6: Solve the system of equations.
Now we have a system of two equations:

1. \( a + 3b = 1 \)
2. \( a - b = 1 \)

We can solve these equations simultaneously. From Equation 2, we can express \( a \) in terms of \( b \):

\[ a = b + 1 \]

### Step 7: Substitute \( a \) in Equation 1.
Substituting \( a \) in Equation 1:

\[ (b + 1) + 3b = 1 \]

This simplifies to:

\[ 4b + 1 = 1 \]

### Step 8: Solve for \( b \).
Subtracting 1 from both sides gives:

\[ 4b = 0 \]

Dividing by 4 yields:

\[ b = 0 \]

### Step 9: Find \( a \) using the value of \( b \).
Now substitute \( b = 0 \) back into Equation 2:

\[ a - 0 = 1 \implies a = 1 \]

### Conclusion:
Thus, the values of \( a \) and \( b \) are:

\[ a = 1, \quad b = 0 \]