Equation of a straight line passing through the point of intersection of `x-y+1=0` and `3x+y-5=0` and perpendicular to one of them is

Options

A. `x+y+3=0`
B. `x+y-3=0`
C. `x-3y+5=0`
D. `x-3y+5=0`

Answer

To find the equation of a straight line that passes through the point of intersection of the lines \(x - y + 1 = 0\) and \(3x + y - 5 = 0\), and is perpendicular to one of them, we will follow these steps:

### Step 1: Find the point of intersection of the two lines.
To find the point of intersection, we will solve the equations simultaneously.

1. The first equation is:
\[ x - y + 1 = 0 \quad \text{(Equation 1)} \]
Rearranging gives:
\[ y = x + 1 \]

2. The second equation is:
\[ 3x + y - 5 = 0 \quad \text{(Equation 2)} \]
Rearranging gives:
\[ y = 5 - 3x \]

Now, we can set the two expressions for \(y\) equal to each other:
\[ x + 1 = 5 - 3x \]
Solving for \(x\):
\[ x + 3x = 5 - 1 \]
\[ 4x = 4 \implies x = 1 \]

Substituting \(x = 1\) back into Equation 1 to find \(y\):
\[ y = 1 + 1 = 2 \]

Thus, the point of intersection is \(P(1, 2)\).

### Step 2: Determine the slope of the line we want to find.
Next, we need to find the slope of one of the given lines to determine the slope of the line that is perpendicular to it.

1. For Equation 1 (\(x - y + 1 = 0\)), we can rewrite it in slope-intercept form \(y = mx + b\):
\[ y = x + 1 \]
Here, the slope \(m_1 = 1\).

2. The slope of the line that is perpendicular to this line is given by:
\[ m = -\frac{1}{m_1} = -\frac{1}{1} = -1 \]

### Step 3: Use the point-slope form to find the equation of the required line.
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1) = (1, 2)\) and \(m = -1\):
\[ y - 2 = -1(x - 1) \]
Expanding this:
\[ y - 2 = -x + 1 \]
Rearranging gives:
\[ x + y - 3 = 0 \]
Thus, the equation of the line passing through the point of intersection and perpendicular to \(x - y + 1 = 0\) is:
\[ x + y - 3 = 0 \]

### Step 4: Verify the equation against the other line.
To check if this line is also perpendicular to the second line \(3x + y - 5 = 0\), we find the slope of the second line.

1. For Equation 2 (\(3x + y - 5 = 0\)):
\[ y = -3x + 5 \]
Here, the slope \(m_2 = -3\).

2. The slope of the line that is perpendicular to this line is:
\[ m = -\frac{1}{m_2} = -\frac{1}{-3} = \frac{1}{3} \]

Since the slope of the line \(x + y - 3 = 0\) is \(-1\), it is not perpendicular to the second line.

### Final Result
The equation of the required line that passes through the point of intersection and is perpendicular to the first line is:
\[ x + y - 3 = 0 \]