Show that the three points (5, 1), (1, -1) and (11, 4) lie on a straight line. Further find
the slope of the line.

Answer

To show that the points (5, 1), (1, -1), and (11, 4) lie on a straight line, we can use the concept of the area of a triangle formed by these three points. If the area is zero, it indicates that the points are collinear.

### Step 1: Identify the points
Let:
- Point A = (5, 1) → (x1, y1)
- Point B = (1, -1) → (x2, y2)
- Point C = (11, 4) → (x3, y3)

### Step 2: Use the area formula
The area of triangle ABC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates of the points:
\[ \text{Area} = \frac{1}{2} \left| 5(-1 - 4) + 1(4 - 1) + 11(1 - (-1)) \right| \]

### Step 3: Simplify the expression
Calculating each term:
- \(5(-1 - 4) = 5 \times -5 = -25\)
- \(1(4 - 1) = 1 \times 3 = 3\)
- \(11(1 - (-1)) = 11 \times 2 = 22\)

Putting it all together:
\[ \text{Area} = \frac{1}{2} \left| -25 + 3 + 22 \right| = \frac{1}{2} \left| 0 \right| = 0 \]

### Step 4: Conclusion about collinearity
Since the area is 0, the points A, B, and C are collinear, meaning they lie on the same straight line.

### Step 5: Find the slope of the line
The slope \(m\) of a line through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
We can use points A and B:
- \(y_1 = 1\), \(y_2 = -1\)
- \(x_1 = 5\), \(x_2 = 1\)

Substituting these values into the slope formula:
\[ m = \frac{-1 - 1}{1 - 5} = \frac{-2}{-4} = \frac{1}{2} \]

### Final Answer
The points (5, 1), (1, -1), and (11, 4) lie on a straight line, and the slope of the line is \(\frac{1}{2}\).

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