Question

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

• A. $L_{1}\left|\right|L_{4}, L_{2}\left|\right|L_{3}, L_{1}$ intersect $L_{4}.$
• B. $L_{1}\,\bot \,L_{2}, L_{1}\left|\right|L_{3}, L_{1}$ intersect $L_{2}.$
• C. $L_{1}\,\bot\, L_{2}, L_{2}\left|\right|L_{3}, L_{1}$ intersect $L_{4}.$
• D. $L_{1}\,\bot\, L_{2}, L_{1}\, \bot\, L_{3}, L_{2}$ intersect $L_{4}.$

Answer 1

Answer: (D)

Consider the lines
$L_{1} : x-y = 1$
$L_{2} : x-y = 1$
$L_{3} : 2x-2y = 5$
$L_{4} : 2x-2y = 7$
$L_{1}\bot L_{2}$ is correct statement
($\because$ Product of their slopes = - 1)
$L_{1}\bot L_{3}$ is also correct statement
($\because$ Product of their slopes $= - 1$)
Now, $L_{2} : x + y = 1$
$L_{4} : 2x - 2y = 7$
$\Rightarrow 2x-2\left(1-x\right) = 7$
$\Rightarrow 2x-2 + 2x = 7$
$\Rightarrow x = \frac{9}{4}$ and $y = \frac{-5}{4}$
Hence, $L_{2}$ intersects $L_{4}.$