Find the equation of the straight line passing through the point (-1,4) and having a gradient of 2:5 .

Options

A. a. 2x-5y+22 =0
B. b. 5x-9y =13
C. c. 13x - 15y +17 =0
D. d. 5x-2y+13 =0

Answer

To find the equation of the straight line passing through the point (-1, 4) with a gradient (slope) of \( \frac{2}{5} \), we can use the point-slope form of the equation of a line. The point-slope form is given by:

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

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

### Step 1: Identify the values
Here, we have:
- \( (x_1, y_1) = (-1, 4) \)
- \( m = \frac{2}{5} \)

### Step 2: Substitute the values into the point-slope formula
Substituting the values into the point-slope formula:

\[ y - 4 = \frac{2}{5}(x - (-1)) \]

This simplifies to:

\[ y - 4 = \frac{2}{5}(x + 1) \]

### Step 3: Distribute the slope
Next, we distribute \( \frac{2}{5} \):

\[ y - 4 = \frac{2}{5}x + \frac{2}{5} \]

### Step 4: Isolate \( y \)
Now, add 4 to both sides to isolate \( y \):

\[ y = \frac{2}{5}x + \frac{2}{5} + 4 \]

To combine the constants, convert 4 to a fraction with a denominator of 5:

\[ 4 = \frac{20}{5} \]

So we have:

\[ y = \frac{2}{5}x + \frac{2}{5} + \frac{20}{5} \]

This simplifies to:

\[ y = \frac{2}{5}x + \frac{22}{5} \]

### Step 5: Rearrange to standard form
To express the equation in standard form \( Ax + By + C = 0 \), we can rearrange the equation:

\[ \frac{2}{5}x - y + \frac{22}{5} = 0 \]

Multiplying through by 5 to eliminate the fraction gives:

\[ 2x - 5y + 22 = 0 \]

### Final Equation
Thus, the equation of the straight line is:

\[ 2x - 5y + 22 = 0 \]