To solve the problem step by step, we will follow the instructions given in the question.
### Step 1: Rewrite the equations
We have two equations to graph:
1. \(3x - 2y = 5\)
2. \(2x = 3y\)
### Step 2: Find points for the first equation \(3x - 2y = 5\)
Let's find two points for this equation by substituting values for \(x\):
- **Point 1**: Let \(x = 1\):
\[
3(1) - 2y = 5 \implies 3 - 2y = 5 \implies -2y = 5 - 3 \implies -2y = 2 \implies y = -1
\]
So, the first point is \((1, -1)\).
- **Point 2**: Let \(x = 3\):
\[
3(3) - 2y = 5 \implies 9 - 2y = 5 \implies -2y = 5 - 9 \implies -2y = -4 \implies y = 2
\]
So, the second point is \((3, 2)\).
### Step 3: Find points for the second equation \(2x = 3y\)
Let's find two points for this equation by substituting values for \(x\):
- **Point 1**: Let \(x = 0\):
\[
2(0) = 3y \implies 0 = 3y \implies y = 0
\]
So, the first point is \((0, 0)\).
- **Point 2**: Let \(x = 3\):
\[
2(3) = 3y \implies 6 = 3y \implies y = 2
\]
So, the second point is \((3, 2)\).
### Step 4: Plot the points on graph paper
Using the scale of \(2 \text{ cm} = 1 \text{ unit}\):
- For the first equation, plot the points \((1, -1)\) and \((3, 2)\).
- For the second equation, plot the points \((0, 0)\) and \((3, 2)\).
### Step 5: Draw the lines
Connect the points for each equation with a straight line:
- Draw a line through \((1, -1)\) and \((3, 2)\) for the first equation.
- Draw a line through \((0, 0)\) and \((3, 2)\) for the second equation.
### Step 6: Find the point of intersection
The lines intersect at the point \((3, 2)\).
### Step 7: Find the area of the triangle formed by the lines and the y-axis
The vertices of the triangle are:
- A: \((0, 0)\) (the origin)
- B: \((0, 2)\) (the y-intercept of the line \(2x = 3y\))
- C: \((3, 2)\) (the intersection point)
To find the area of triangle ABC:
- Base = length along the y-axis from (0, 0) to (0, 2) = 2 units
- Height = length along the x-axis from (0, 0) to (3, 0) = 3 units
Using the formula for the area of a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 3 = 3 \text{ square units}
\]
### Final Answer
- The coordinates of the point of intersection are \((3, 2)\).
- The area of the triangle formed by the lines and the y-axis is \(3 \text{ square units}\).
