Question

The equation of the lines through \( (1, 1) \) and making angles of \( 45^\circ \) with the line \( x + y = 0 \) are:

• A. \( x-1=0, x-y=0 \)
• B. \( x-y=0, y-1=0 \)
• C. \( x+y-2=0, y-1=0 \)
• D. \( x-1=0, y-1=0 \)

Hint:

The line \( x+y=0 \) has an inclination of \( 135^\circ \) with the positive x-axis. Lines making an angle of \( 45^\circ \) with it must have inclinations:
\( 135^\circ - 45^\circ = 90^\circ \) (vertical line)
\( 135^\circ + 45^\circ = 180^\circ \) (horizontal line)
The vertical and horizontal lines passing through \( (1, 1) \) are simply \( x=1 \) and \( y=1 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the equations of two straight lines that pass through the point \( (1, 1) \) and make an angle of \( 45^\circ \) with the line \( x + y = 0 \).

Step 2: Key Formula or Approach:
The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \]

Step 3: Detailed Explanation:
The given line is \( x + y = 0 \implies y = -x \).
The slope of this line is \( m_1 = -1 \).
Let the slope of the required line be \( m \). Given \( \theta = 45^\circ \):
\[ \tan 45^\circ = \left| \frac{m - (-1)}{1 + m(-1)} \right| \implies 1 = \left| \frac{m + 1}{1 - m} \right| \] This gives two cases:

Case 1:
\[ \frac{m + 1}{1 - m} = 1 \implies m + 1 = 1 - m \implies 2m = 0 \implies m = 0 \] The line with slope \( m = 0 \) passing through \( (1, 1) \) is:
\[ y - 1 = 0(x - 1) \implies y - 1 = 0 \]

Case 2:
\[ \frac{m + 1}{1 - m} = -1 \implies m + 1 = m - 1 \implies 1 = -1 \, \text{(No real solution)} \] This indicates that the second line is vertical, having an undefined slope (\( m \to \infty \)).
The vertical line passing through \( (1, 1) \) is:
\[ x - 1 = 0 \] Thus, the two required lines are \( x - 1 = 0 \) and \( y - 1 = 0 \).

Step 4: Final Answer:
(D) \( x-1=0, y-1=0 \)