The eq. of the straight line which passes through the point (1,-2) and cuts of equal intercepts from the axes will be

Options

A. `x+y=1`
B. `x-y=1`
C. `x+y+1=0`
D. `x-y-3=0`

Answer

To find the equation of the straight line that passes through the point (1, -2) and cuts equal intercepts from the axes, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Condition of Equal Intercepts:**
The line cuts equal intercepts on the x-axis and y-axis. Let's denote the intercepts as \( A \) (on the x-axis) and \( B \) (on the y-axis). Since they are equal, we can set \( A = B = c \).

2. **Equation of the Line in Intercept Form:**
The equation of a line in intercept form is given by:
\[ \frac{x}{A} + \frac{y}{B} = 1 \]
Since \( A = B = c \), we can rewrite the equation as:
\[ \frac{x}{c} + \frac{y}{c} = 1 \]
This simplifies to:
\[ x + y = c \]

3. **Substituting the Point (1, -2):**
We know that the line passes through the point (1, -2). We can substitute \( x = 1 \) and \( y = -2 \) into the equation:
\[ 1 + (-2) = c \]
This simplifies to:
\[ -1 = c \]

4. **Final Equation of the Line:**
Now that we have \( c = -1 \), we can substitute this back into the equation of the line:
\[ x + y = -1 \]
Rearranging gives us:
\[ x + y + 1 = 0 \]

5. **Conclusion:**
The equation of the straight line that passes through the point (1, -2) and cuts equal intercepts from the axes is:
\[ x + y + 1 = 0 \]