Two forces `F_(1)={2, 3} and F_(2)={4, 1}` are specified relative to a general cartesian form. Their points of application are respectivel, A=(1, 1) and B=(2, 4). Find the coordinates of the resultant and the equation of the straight line l containing it.

Answer

To solve the problem step by step, we will find the resultant of the two forces and then determine the equation of the straight line containing the points of application of these forces.

### Step 1: Represent the Forces in Vector Form
The forces are given as:
- \( F_1 = \{2, 3\} \)
- \( F_2 = \{4, 1\} \)

In vector form, these can be represented as:
- \( F_1 = 2\mathbf{i} + 3\mathbf{j} \)
- \( F_2 = 4\mathbf{i} + 1\mathbf{j} \)

### Step 2: Find the Resultant Force
The resultant force \( F_R \) is the vector sum of \( F_1 \) and \( F_2 \):
\[ F_R = F_1 + F_2 = (2\mathbf{i} + 3\mathbf{j}) + (4\mathbf{i} + 1\mathbf{j}) \]
Combining the components:
\[ F_R = (2 + 4)\mathbf{i} + (3 + 1)\mathbf{j} = 6\mathbf{i} + 4\mathbf{j} \]

### Step 3: Identify the Points of Application
The points of application of the forces are given as:
- Point A: \( (1, 1) \)
- Point B: \( (2, 4) \)

### Step 4: Find the Equation of the Line Containing Points A and B
To find the equation of the line through points A and B, we first need to calculate the slope \( m \):
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 1}{2 - 1} = \frac{3}{1} = 3 \]

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
We can use point A \( (1, 1) \):
\[ y - 1 = 3(x - 1) \]
Expanding this:
\[ y - 1 = 3x - 3 \]
Rearranging gives:
\[ 3x - y - 2 = 0 \]
This can also be written as:
\[ 3x - y = 2 \]

### Final Answers
- The coordinates of the resultant force are \( F_R = 6\mathbf{i} + 4\mathbf{j} \).
- The equation of the straight line containing points A and B is \( 3x - y = 2 \).