Area of triangle formed by straight line `y = mx + c` with coordinate area

Options

A. `(c ^(2))/( m )`
B. `( m )/(c ^(2))`
C. `(c ^(2))/( 2m)`
D. `( 2m )/(c ^(2))`

Answer

To find the area of the triangle formed by the straight line \( y = mx + c \) with the coordinate axes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Line Equation**:
The equation of the line is given as:
\[ y = mx + c \]

2. **Find the Intercepts**:
- **Y-Intercept**: Set \( x = 0 \) in the line equation:
\[ y = m(0) + c = c \]
So, the y-intercept is \( (0, c) \).

- **X-Intercept**: Set \( y = 0 \) in the line equation:
\[ 0 = mx + c \implies mx = -c \implies x = -\frac{c}{m} \]
So, the x-intercept is \( \left(-\frac{c}{m}, 0\right) \).

3. **Plot the Triangle**:
The triangle is formed by the line and the coordinate axes. The vertices of the triangle are:
- \( (0, 0) \) (origin)
- \( (0, c) \) (y-intercept)
- \( \left(-\frac{c}{m}, 0\right) \) (x-intercept)

4. **Calculate the Area of the Triangle**:
The area \( A \) of a triangle formed by the intercepts can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the length along the x-axis from the origin to the x-intercept, which is \( \left| -\frac{c}{m} \right| = \frac{c}{m} \), and the height is the y-intercept \( c \).

Therefore, the area is:
\[ A = \frac{1}{2} \times \frac{c}{m} \times c = \frac{c^2}{2m} \]

5. **Conclusion**:
The area of the triangle formed by the line \( y = mx + c \) with the coordinate axes is:
\[ \frac{c^2}{2m} \]