Draw the graphs of the following linear equations on the
same graph paper:
2x + 3y = 12, x − y = 1
Find the coordinates of the vertices of the triangle formed by the two
straight lines and the area bounded by these lines and x-axis

To address this problem, let's proceed step-by-step:

### Step 1: Convert the Equations to Slope-Intercept Form

We begin by converting the given linear equations into slope-intercept form \((y = mx + c)\).

1. **Equation 1: \(2x + 3y = 12\)**
   \[
   3y = -2x + 12
   \]
   \[
   y = -\frac{2}{3}x + 4
   \]

2. **Equation 2: \(x - y = 1\)**
   \[
   y = x - 1
   \]

### Step 2: Plot the Equations

Now, let's plot these two lines on the same graph.

- **Line 1: \(y = -\frac{2}{3}x + 4\)**
  - **Intercepts**: \(x\)-intercept when \(y = 0\):
    \[
    0 = -\frac{2}{3}x + 4 \Rightarrow x = 6
    \]
    So, \( (6, 0) \).
  - **\(y\)-intercept**: \(x = 0\),
    \[
    y = 4
    \]
    So, \( (0, 4) \).

- **Line 2: \(y = x - 1\)**
  - **Intercepts**: \(x\)-intercept when \(y = 0\):
    \[
    0 = x - 1 \Rightarrow x = 1
    \]
    So, \( (1, 0) \).
  - **\(y\)-intercept**: \(x = 0\),
    \[
    y = -1
    \]
    So, \( (0, -1) \).

### Step 3: Find the Intersection Point of the Two Lines

To find the coordinates of the intersection, solve the system of equations:

\[
\begin{aligned}
y &= -\frac{2}{3}x + 4, \\
y &= x - 1.
\end{aligned}
\]
Set the equations equal to each other:

\[
-\frac{2}{3}x + 4 = x - 1
\]
Multiply by 3 to eliminate the fraction:

\[
-2x + 12 = 3x - 3
\]
\[
15 = 5x
\]
\[
x = 3
\]
Substitute \(x = 3\) into one of the original equations:

\[
y = 3 - 1 = 2
\]
So, the intersection point is \((3, 2)\).

### Step 4: Determine the Vertices of the Triangle

The vertices of the triangle are where the lines intersect the \(x\)-axis and the intersection point itself:
- Intersection of Line 1 with \(x\)-axis: \((6, 0)\),
- Intersection of Line 2 with \(x\)-axis: \((1, 0)\),
- Intersection of both lines: \((3, 2)\).

Thus, the vertices are \((6, 0)\), \((1, 0)\), and \((3, 2)\).

### Step 5: Calculate the Area of the Triangle

The area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:

\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the vertices:

\[
\text{Area} = \frac{1}{2} \left| 6(0 - 2) + 1(2 - 0) + 3(0 - 0) \right|
\]
\[
= \frac{1}{2} \left| -12 + 2 \right|
\]
\[
= \frac{1}{2} \times 10 = 5 \text{ square units}
\]

### Final Result

The area of the triangle is 5 square units. The vertices of the triangle are \((6, 0)\), \((1, 0)\), and \((3, 2)\).

Would you like a visual representation of this graph?

Learn how to graph the linear equations 2x + 3y = 12 and x - y = 1 on the same graph paper. Find the coordinates of the vertices of the triangle formed by these lines and calculate the area bounded by these lines and the x-axis. Suitable for high school students studying linear equations and triangle geometry.

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