In each of the following find the value of 'k', for which the points are colinear.
(7,-2), (5,1), (3,k)

Answer

To find the value of 'k' for which the points (7, -2), (5, 1), and (3, k) are collinear, we can use the concept of the area of a triangle formed by these three points. If the points are collinear, the area of the triangle formed by them will be zero.

### Step 1: Set up the area formula
The area \( A \) of a triangle formed by three points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:

\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 2: Substitute the points into the formula
For our points \((7, -2)\), \((5, 1)\), and \((3, k)\), we substitute:

- \(x_1 = 7\), \(y_1 = -2\)
- \(x_2 = 5\), \(y_2 = 1\)
- \(x_3 = 3\), \(y_3 = k\)

Substituting these values into the area formula gives:

\[ A = \frac{1}{2} \left| 7(1 - k) + 5(k + 2) + 3(-2 - 1) \right| \]

### Step 3: Simplify the expression
Now, we simplify the expression inside the absolute value:

\[ A = \frac{1}{2} \left| 7(1 - k) + 5(k + 2) + 3(-3) \right| \]

Calculating each term:

- \(7(1 - k) = 7 - 7k\)
- \(5(k + 2) = 5k + 10\)
- \(3(-3) = -9\)

Combining these gives:

\[ A = \frac{1}{2} \left| 7 - 7k + 5k + 10 - 9 \right| \]

This simplifies to:

\[ A = \frac{1}{2} \left| -2k + 8 \right| \]

### Step 4: Set the area equal to zero
Since the points are collinear, the area \( A \) must be zero:

\[ \frac{1}{2} \left| -2k + 8 \right| = 0 \]

This implies:

\[ \left| -2k + 8 \right| = 0 \]

### Step 5: Solve for 'k'
The absolute value equation gives us:

\[ -2k + 8 = 0 \]

Solving for \( k \):

\[ -2k = -8 \\ k = 4 \]

### Final Answer
The value of \( k \) for which the points (7, -2), (5, 1), and (3, k) are collinear is \( k = 4 \).

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