If three lines `3x+4y=12," "5x+8y=40` and x axis form a triangle. Find its area

Answer

To find the area of the triangle formed by the lines \(3x + 4y = 12\), \(5x + 8y = 40\), and the x-axis, we can follow these steps:

### Step 1: Find the x-intercepts of the lines
To find the x-intercepts, we set \(y = 0\) in both equations.

1. For the first line \(3x + 4y = 12\):
\[ 3x + 4(0) = 12 \implies 3x = 12 \implies x = 4 \]
So, the x-intercept is \((4, 0)\).

2. For the second line \(5x + 8y = 40\):
\[ 5x + 8(0) = 40 \implies 5x = 40 \implies x = 8 \]
So, the x-intercept is \((8, 0)\).

### Step 2: Find the intersection point of the two lines
To find the intersection point of the lines, we will solve the equations simultaneously.

1. The equations are:
\[ 3x + 4y = 12 \quad \text{(1)} \]
\[ 5x + 8y = 40 \quad \text{(2)} \]

2. To eliminate \(x\), we can multiply equation (1) by 5 and equation (2) by 3:
\[ 15x + 20y = 60 \quad \text{(3)} \]
\[ 15x + 24y = 120 \quad \text{(4)} \]

3. Now, subtract equation (3) from equation (4):
\[ (15x + 24y) - (15x + 20y) = 120 - 60 \]
\[ 4y = 60 \implies y = 15 \]

4. Substitute \(y = 15\) back into equation (1) to find \(x\):
\[ 3x + 4(15) = 12 \implies 3x + 60 = 12 \implies 3x = 12 - 60 \implies 3x = -48 \implies x = -16 \]
So, the intersection point is \((-16, 15)\).

### Step 3: Calculate the area of the triangle
The area \(A\) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]

1. The base of the triangle is the distance between the x-intercepts:
\[ \text{Base} = 8 - 4 = 4 \]

2. The height of the triangle is the y-coordinate of the intersection point:
\[ \text{Height} = 15 \]

3. Now, substitute the values into the area formula:
\[ A = \frac{1}{2} \times 4 \times 15 = \frac{60}{2} = 30 \]

Thus, the area of the triangle is \(30\) square units.