Find the equation of a straight line passing through (2, -3) and having a slope of 1 unit.

Options

A. y - x + 5 = 0
B. x + y + 5 = 0
C. - x - y = -5
D. x - y - 5 = 0

Answer

To find the equation of a straight line passing through the point (2, -3) with a slope of 1, 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:
- \( (x_1, y_1) \) is a point on the line,
- \( m \) is the slope of the line.

### Step 1: Identify the given values
Here, we have:
- Point \( (x_1, y_1) = (2, -3) \)
- Slope \( m = 1 \)

### Step 2: Substitute the values into the point-slope form
Substituting the values into the point-slope form, we get:

\[ y - (-3) = 1(x - 2) \]

### Step 3: Simplify the equation
This simplifies to:

\[ y + 3 = 1(x - 2) \]

### Step 4: Distribute the slope
Now, distribute the slope on the right side:

\[ y + 3 = x - 2 \]

### Step 5: Isolate \( y \)
To isolate \( y \), subtract 3 from both sides:

\[ y = x - 2 - 3 \]

This simplifies to:

\[ y = x - 5 \]

### Step 6: Write in standard form (optional)
If we want to write this in standard form, we can rearrange it:

\[ x - y - 5 = 0 \]

Thus, the equation of the straight line is:

\[ x - y - 5 = 0 \]

### Final Answer:
The equation of the line is \( x - y - 5 = 0 \).

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