Equation of a line passing through the point `(2,3)` and perpendicular to the line `x+y+1=0` is :

Options

A. `y-x+1=0`
B. `x-y+1=0`
C. `x+y-1=0`
D. None of these

Answer

To find the equation of a line that passes through the point (2, 3) and is perpendicular to the line given by the equation \(x + y + 1 = 0\), we can follow these steps:

### Step 1: Determine the slope of the given line
The equation of the line can be rewritten in slope-intercept form \(y = mx + b\). Starting from:
\[ x + y + 1 = 0 \]
we can isolate \(y\):
\[ y = -x - 1 \]
From this, we can see that the slope \(m_1\) of the given line is \(-1\).

**Hint:** To find the slope from the standard form \(Ax + By + C = 0\), rearrange it to the form \(y = mx + b\).

### Step 2: Find the slope of the perpendicular line
The slope of a line that is perpendicular to another line is the negative reciprocal of the original line's slope. Therefore, if the slope of the given line is \(-1\), the slope \(m_2\) of the perpendicular line is:
\[ m_2 = -\frac{1}{-1} = 1 \]

**Hint:** The negative reciprocal of a slope \(m\) is given by \(-\frac{1}{m}\).

### Step 3: Use the point-slope form of the equation of a line
The point-slope form of the equation of a line is given by:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1)\) is a point on the line and \(m\) is the slope. Here, we have the point \((2, 3)\) and the slope \(1\):
\[ y - 3 = 1(x - 2) \]

**Hint:** The point-slope form is useful when you know a point on the line and the slope.

### Step 4: Simplify the equation
Now, we simplify the equation:
\[ y - 3 = x - 2 \]
Adding \(3\) to both sides gives:
\[ y = x + 1 \]

**Hint:** Always simplify the equation to its most basic form to easily identify the slope and y-intercept.

### Step 5: Convert to standard form
To express the equation in standard form \(Ax + By + C = 0\), we rearrange:
\[ x - y + 1 = 0 \]

**Hint:** Standard form is often preferred in mathematics, where \(A\), \(B\), and \(C\) are integers.

### Final Answer
The equation of the line passing through the point \((2, 3)\) and perpendicular to the line \(x + y + 1 = 0\) is:
\[ x - y + 1 = 0 \]