What is the equation of the line if slope is ` - (2)/(5)` and it passes through the point` (1,-3)` ?

Options

A. ` 2 x + 5y = 17`
B. ` 2 x - 5y = - 13`
C. ` 2 x - 5y = 17`
D. ` 2 x + 5y = - 13`

Answer

To find the equation of the line with a given slope and a point it passes through, we can use the point-slope form of the equation of a line, which is given by:

\[ y - y_1 = m(x - x_1) \]

where:
- \( m \) is the slope of the line,
- \( (x_1, y_1) \) is a point on the line.

### Step-by-Step Solution:

1. **Identify the given values:**
- Slope \( m = -\frac{2}{5} \)
- Point \( (x_1, y_1) = (1, -3) \)

2. **Substitute the values into the point-slope form:**
\[ y - (-3) = -\frac{2}{5}(x - 1) \]
This simplifies to:
\[ y + 3 = -\frac{2}{5}(x - 1) \]

3. **Distribute the slope on the right side:**
\[ y + 3 = -\frac{2}{5}x + \frac{2}{5} \]

4. **Isolate \( y \) by subtracting 3 from both sides:**
\[ y = -\frac{2}{5}x + \frac{2}{5} - 3 \]

5. **Convert -3 to a fraction with a denominator of 5:**
\[ -3 = -\frac{15}{5} \]
So, we have:
\[ y = -\frac{2}{5}x + \frac{2}{5} - \frac{15}{5} \]

6. **Combine the constant terms:**
\[ y = -\frac{2}{5}x - \frac{13}{5} \]

7. **Rearranging to standard form \( Ax + By + C = 0 \):**
\[ \frac{2}{5}x + y + \frac{13}{5} = 0 \]
To eliminate the fraction, multiply through by 5:
\[ 2x + 5y + 13 = 0 \]

8. **Final equation of the line:**
\[ 2x + 5y + 13 = 0 \]