Solve graphically the pair of linear equations `3x – 4y + 3 = 0` and `3x + 4y – 21 =0` Find the co-ordinates of vertices of triangular region formed by these lines and x-axis. Also calculate the area of this triangle.

Answer

To solve the pair of linear equations graphically and find the area of the triangle formed by the lines and the x-axis, we will follow these steps:

### Step 1: Write the equations in slope-intercept form
The given equations are:
1. \(3x - 4y + 3 = 0\)
2. \(3x + 4y - 21 = 0\)

We can rearrange these equations to find the y-intercept.

**For the first equation:**
\[ 3x - 4y + 3 = 0 \implies 4y = 3x + 3 \implies y = \frac{3}{4}x + \frac{3}{4} \]

**For the second equation:**
\[ 3x + 4y - 21 = 0 \implies 4y = -3x + 21 \implies y = -\frac{3}{4}x + \frac{21}{4} \]

### Step 2: Find points to plot the lines
To graph these equations, we need at least two points for each line.

**For the first equation \(y = \frac{3}{4}x + \frac{3}{4}\):**
- Let \(y = 0\):
\[ 0 = \frac{3}{4}x + \frac{3}{4} \implies \frac{3}{4}x = -\frac{3}{4} \implies x = -1 \quad \text{(Point: } (-1, 0)\text{)} \]
- Let \(x = 3\):
\[ y = \frac{3}{4}(3) + \frac{3}{4} = \frac{9}{4} + \frac{3}{4} = 3 \quad \text{(Point: } (3, 3)\text{)} \]

**For the second equation \(y = -\frac{3}{4}x + \frac{21}{4}\):**
- Let \(y = 0\):
\[ 0 = -\frac{3}{4}x + \frac{21}{4} \implies \frac{3}{4}x = \frac{21}{4} \implies x = 7 \quad \text{(Point: } (7, 0)\text{)} \]
- Let \(x = 3\):
\[ y = -\frac{3}{4}(3) + \frac{21}{4} = -\frac{9}{4} + \frac{21}{4} = 3 \quad \text{(Point: } (3, 3)\text{)} \]

### Step 3: Plot the points and draw the lines
Now we have the points:
- From the first equation: \((-1, 0)\) and \((3, 3)\)
- From the second equation: \((7, 0)\) and \((3, 3)\)

Plot these points on a graph and draw the lines. The lines will intersect at the point \((3, 3)\).

### Step 4: Identify the vertices of the triangular region
The vertices of the triangle formed by the lines and the x-axis are:
1. \((-1, 0)\)
2. \((7, 0)\)
3. \((3, 3)\)

### Step 5: Calculate the area of the triangle
To find the area of the triangle formed by these points, we can use the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

**Base:** The distance between points \((-1, 0)\) and \((7, 0)\):
\[ \text{Base} = 7 - (-1) = 8 \]

**Height:** The height is the y-coordinate of the point \((3, 3)\):
\[ \text{Height} = 3 \]

Now substituting these values into the area formula:
\[ \text{Area} = \frac{1}{2} \times 8 \times 3 = 12 \text{ square units} \]

### Final Answer
The coordinates of the vertices of the triangular region are:
1. \((-1, 0)\)
2. \((7, 0)\)
3. \((3, 3)\)

The area of the triangle is \(12\) square units.

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