The joint equation of the straight lines `x+y=1` and `x-y=4`, is

Options

A. `x^(2)-y^(2)=-4`
B. `x^(2)-y^(2)=4`
C. `(x+y-1)(x-y-4)=0`
D. `(x+y+1)(x-y+4)=0`

Answer

To find the joint equation of the straight lines given by the equations \(x + y = 1\) and \(x - y = 4\), we can follow these steps:

### Step 1: Write the equations in standard form
We start with the equations of the lines:
1. \(x + y = 1\) can be rewritten as:
\[ x + y - 1 = 0 \]
2. \(x - y = 4\) can be rewritten as:
\[ x - y - 4 = 0 \]

### Step 2: Multiply the two equations
To find the joint equation, we multiply the left-hand sides (LHS) of the two equations:
\[ (x + y - 1)(x - y - 4) = 0 \]

### Step 3: Expand the product
Now we will expand the product:
\[ (x + y - 1)(x - y - 4) = x(x - y - 4) + y(x - y - 4) - 1(x - y - 4) \]
Expanding each term:
1. \(x(x - y - 4) = x^2 - xy - 4x\)
2. \(y(x - y - 4) = xy - y^2 - 4y\)
3. \(-1(x - y - 4) = -x + y + 4\)

Combining these, we get:
\[ x^2 - xy - 4x + xy - y^2 - 4y - x + y + 4 = 0 \]

### Step 4: Combine like terms
Now we combine like terms:
- The \(xy\) terms cancel out:
\[ x^2 - y^2 - 5x - 3y + 4 = 0 \]

### Final Result
Thus, the joint equation of the straight lines \(x + y = 1\) and \(x - y = 4\) is:
\[ x^2 - y^2 - 5x - 3y + 4 = 0 \]