Consider the straight lines $$L_1 : x - y = 1$$, $$L_2 : x + y = 1$$, $$L_3 : 2x + 2y = 5$$, $$L_4 : 2x - 2y = 7$$. The correct statement is

First, write each line in the slope-intercept form $$y = mx + c$$ to read its slope quickly.

$$L_1:\;x - y = 1 \;\Longrightarrow\; y = x - 1 \;\; \Rightarrow\; m_1 = 1$$

$$L_2:\;x + y = 1 \;\Longrightarrow\; y = -x + 1 \;\Rightarrow\; m_2 = -1$$

$$L_3:\;2x + 2y = 5 \;\Longrightarrow\; x + y = \tfrac{5}{2} \;\Longrightarrow\; y = -x + \tfrac{5}{2} \;\Rightarrow\; m_3 = -1$$

$$L_4:\;2x - 2y = 7 \;\Longrightarrow\; x - y = \tfrac{7}{2} \;\Longrightarrow\; y = x - \tfrac{7}{2} \;\Rightarrow\; m_4 = 1$$

Now apply the two standard tests.

1. Parallel lines have equal slopes: $$m_p = m_q$$.
2. Perpendicular lines satisfy $$m_p \, m_q = -1$$.

$$m_1 \cdot m_2 = 1 \times (-1) = -1 \;\Rightarrow\; L_1 \perp L_2$$

$$m_1 \cdot m_3 = 1 \times (-1) = -1 \;\Rightarrow\; L_1 \perp L_3$$

$$m_2 = m_3 = -1 \;\Rightarrow\; L_2 \parallel L_3$$

$$m_1 = m_4 = 1 \;\Rightarrow\; L_1 \parallel L_4$$

$$m_2 \neq m_4 \quad$$ and $$\;m_2 \, m_4 = (-1)\times 1 = -1 \neq 1$$, so the slopes are different; therefore the two lines are not parallel, hence they meet. (Their product being $$-1$$ also shows they are perpendicular.) Thus $$L_2$$ intersects $$L_4$$.

Summarising:

• $$L_1 \perp L_2$$
• $$L_1 \perp L_3$$
• $$L_2$$ intersects $$L_4$$ (they are perpendicular).
• Any other claimed relations in the options are incorrect (e.g., $$L_1$$ does not intersect $$L_4$$ because they are parallel).

Only Option D lists all three correct relations.

Final Answer: Option D which is: $$L_1 \perp L_2, \; L_1 \perp L_3, \; L_2$$ intersect $$L_4$$.