Q56 / 90 · JEE Main 2012 (26 May Online) · Mathematics
Consider the straight lines
$$
\begin{aligned}
& L_1: x-y=1 \\
& L_2: x+y=1 \\
& L_3: 2 x+2 y=5 \\
& L_4: 2 x-2 y=7
\end{aligned}
$$
The correct statement is
- A$L_1\left\|L_4, L_2\right\| L_3, L_1$ intersect $L_4$.
- B$L_1 \perp L_2, L_1 \| L_3, L_1$ intersect $L_2$.
- C$L_1 \perp L_2, L_2 \| L_3, L_1$ intersect $L_4$.
- Correct answerD$L_1 \perp L_2, L_1 \perp L_3, L_2$ intersect $L_4$.
Solution
Consider the lines
$L_1: x-y=1$
$L_2: x+\mathrm{y}=1$
$L_3: 2 x+2 y=5$
$L_4: 2 x-2 y=7$
$L_1 \perp L_2$ is correct statement
$(\because$ Product of their slopes $=-1)$
$L_1 \perp L_3$ is also correct statement
$(\because$ Product of their slopes $=-1)$
Now, $L_2: x+y=1$
$L_4: 2 x-2 y=7$
$\Rightarrow 2 x-2(1-x)=7$
$\Rightarrow 2 x-2+2 x=7$
$\Rightarrow x=\frac{9}{4}$ and $y=\frac{-5}{4}$
Hence, $L_2$ intersects $L_4$.
Hence, the correct answer is option: 'D'.