Find the graphically, the vertices of the triangle whose sides have the equations `2y-x=8, 5y-x=14` and `y-2x=1` respectively.
Take 1 cm=1 unit on both the axes.

Answer

To find the vertices of the triangle formed by the lines given by the equations \(2y - x = 8\), \(5y - x = 14\), and \(y - 2x = 1\) graphically, we will follow these steps:

### Step 1: Convert the equations to slope-intercept form
We will convert each equation into the form \(y = mx + b\) where \(m\) is the slope and \(b\) is the y-intercept.

1. **For the first equation**:
\[ 2y - x = 8 \implies 2y = x + 8 \implies y = \frac{1}{2}x + 4 \]

2. **For the second equation**:
\[ 5y - x = 14 \implies 5y = x + 14 \implies y = \frac{1}{5}x + \frac{14}{5} \]

3. **For the third equation**:
\[ y - 2x = 1 \implies y = 2x + 1 \]

### Step 2: Find points for each line
Next, we will find points for each line by substituting values for \(x\) and calculating \(y\).

1. **For \(y = \frac{1}{2}x + 4\)**:
- If \(x = 0\), \(y = 4\) → Point (0, 4)
- If \(x = -2\), \(y = 3\) → Point (-2, 3)
- If \(x = 4\), \(y = 6\) → Point (4, 6)

2. **For \(y = \frac{1}{5}x + \frac{14}{5}\)**:
- If \(x = 0\), \(y = \frac{14}{5} = 2.8\) → Point (0, 2.8)
- If \(x = 5\), \(y = 4\) → Point (5, 4)
- If \(x = -4\), \(y = 2\) → Point (-4, 2)

3. **For \(y = 2x + 1\)**:
- If \(x = 0\), \(y = 1\) → Point (0, 1)
- If \(x = 1\), \(y = 3\) → Point (1, 3)
- If \(x = 2\), \(y = 5\) → Point (2, 5)

### Step 3: Plot the points on a graph
Using the points calculated, we will plot them on a graph with both axes scaled such that 1 cm = 1 unit.

### Step 4: Draw the lines
Connect the points for each equation to form the lines:
- Line 1: Connect points (0, 4), (-2, 3), and (4, 6).
- Line 2: Connect points (0, 2.8), (5, 4), and (-4, 2).
- Line 3: Connect points (0, 1), (1, 3), and (2, 5).

### Step 5: Identify the intersection points
The vertices of the triangle will be the intersection points of the lines:
1. **Intersection of Line 1 and Line 2**: Solve \(2y - x = 8\) and \(5y - x = 14\).
2. **Intersection of Line 2 and Line 3**: Solve \(5y - x = 14\) and \(y - 2x = 1\).
3. **Intersection of Line 1 and Line 3**: Solve \(2y - x = 8\) and \(y - 2x = 1\).

### Step 6: Calculate the intersection points
1. **For Line 1 and Line 2**:
\[ 2y - x = 8 \quad (1) \\ 5y - x = 14 \quad (2) \]
Subtract (1) from (2):
\[ 3y = 6 \implies y = 2 \\ \text{Substituting } y = 2 \text{ in (1): } 2(2) - x = 8 \implies x = -4 \\ \text{Intersection point: } (-4, 2) \]

2. **For Line 2 and Line 3**:
\[ 5y - x = 14 \quad (1) \\ y - 2x = 1 \quad (2) \]
Rearranging (2):
\[ y = 2x + 1 \\ \text{Substituting in (1): } 5(2x + 1) - x = 14 \implies 10x + 5 - x = 14 \implies 9x = 9 \implies x = 1 \\ \text{Substituting } x = 1 \text{ in (2): } y = 2(1) + 1 = 3 \\ \text{Intersection point: } (1, 3) \]

3. **For Line 1 and Line 3**:
\[ 2y - x = 8 \quad (1) \\ y - 2x = 1 \quad (2) \]
Rearranging (2):
\[ y = 2x + 1 \\ \text{Substituting in (1): } 2(2x + 1) - x = 8 \implies 4x + 2 - x = 8 \implies 3x = 6 \implies x = 2 \\ \text{Substituting } x = 2 \text{ in (2): } y = 2(2) + 1 = 5 \\ \text{Intersection point: } (2, 5) \]

### Conclusion
The vertices of the triangle formed by the given lines are:
1. \((-4, 2)\)
2. \((1, 3)\)
3. \((2, 5)\)